package Algorithms.tree;
import java.util.ArrayList;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.Stack;
/**
* REFS:
* http://blog.csdn.net/fightforyourdream/article/details/16843303 面试大总结之二：Java搞定面试中的二叉树题目
* http://blog.csdn.net/luckyxiaoqiang/article/details/7518888 轻松搞定面试中的二叉树题目
* http://www.cnblogs.com/Jax/archive/2009/12/28/1633691.html 算法大全（3） 二叉树
*
* 1. 求二叉树中的节点个数: getNodeNumRec（递归），getNodeNum（迭代）
* 2. 求二叉树的深度: getDepthRec（递归），getDepth
* 3. 前序遍历，中序遍历，后序遍历: preorderTraversalRec, preorderTraversal, inorderTraversalRec, postorderTraversalRec
* (https://en.wikipedia.org/wiki/Tree_traversal#Pre-order_2)
* 4.分层遍历二叉树（按层次从上往下，从左往右）: levelTraversal, levelTraversalRec（递归解法）
* 5. 将二叉查找树变为有序的双向链表: convertBST2DLLRec, convertBST2DLL
* 6. 求二叉树第K层的节点个数：getNodeNumKthLevelRec, getNodeNumKthLevel
* 7. 求二叉树中叶子节点的个数：getNodeNumLeafRec, getNodeNumLeaf
* 8. 判断两棵二叉树是否相同的树：isSameRec, isSame
* 9. 判断二叉树是不是平衡二叉树：isAVLRec
* 10. 求二叉树的镜像（破坏和不破坏原来的树两种情况）：
* mirrorRec, mirrorCopyRec
* mirror, mirrorCopy
* 10.1 判断两个树是否互相镜像：isMirrorRec isMirror
* 11. 求二叉树中两个节点的最低公共祖先节点：
* LAC 求解最小公共祖先, 使用list来存储path.
* LCABstRec 递归求解BST树.
* LCARec 递归算法 .
* 12. 求二叉树中节点的最大距离：getMaxDistanceRec
* 13. 由前序遍历序列和中序遍历序列重建二叉树：rebuildBinaryTreeRec
* 14. 判断二叉树是不是完全二叉树：isCompleteBinaryTree, isCompleteBinaryTreeRec
* 15. 找出二叉树中最长连续子串(即全部往左的连续节点，或是全部往右的连续节点）findLongest
*/
public class TreeDemo {
/*
1
/ \
2 3
/ \ \
4 5 6
*/
public static void main(String[] args) {
TreeNode r1 = new TreeNode(1);
TreeNode r2 = new TreeNode(2);
TreeNode r3 = new TreeNode(3);
TreeNode r4 = new TreeNode(4);
TreeNode r5 = new TreeNode(5);
TreeNode r6 = new TreeNode(6);
/*
10
/ \
6 14
/ \ \
4 8 16
/
0
*/
/*
1
/ \
2 3
/ \ \
4 5 6
*/
// TreeNode r1 = new TreeNode(10);
// TreeNode r2 = new TreeNode(6);
// TreeNode r3 = new TreeNode(14);
// TreeNode r4 = new TreeNode(4);
// TreeNode r5 = new TreeNode(8);
// TreeNode r6 = new TreeNode(16);
TreeNode r7 = new TreeNode(0);
r1.left = r2;
r1.right = r3;
r2.left = r4;
r2.right = r5;
r3.right = r6;
r4.left = r7;
TreeNode t1 = new TreeNode(10);
TreeNode t2 = new TreeNode(6);
TreeNode t3 = new TreeNode(14);
TreeNode t4 = new TreeNode(4);
TreeNode t5 = new TreeNode(8);
TreeNode t6 = new TreeNode(16);
TreeNode t7 = new TreeNode(0);
TreeNode t8 = new TreeNode(0);
TreeNode t9 = new TreeNode(0);
TreeNode t10 = new TreeNode(0);
TreeNode t11 = new TreeNode(0);
t1.left = t2;
t1.right = t3;
t2.left = t4;
t2.right = t5;
t3.left = t6;
t3.right = t7;
t4.left = t8;
//t4.right = t9;
t5.right = t9;
// test distance
// t5.right = t8;
// t8.right = t9;
// t9.right = t10;
// t10.right = t11;
/*
10
/ \
6 14
/ \ \
4 8 16
/
0
*/
// System.out.println(LCABstRec(t1, t2, t4).val);
// System.out.println(LCABstRec(t1, t2, t6).val);
// System.out.println(LCABstRec(t1, t4, t6).val);
// System.out.println(LCABstRec(t1, t4, t7).val);
// System.out.println(LCABstRec(t1, t3, t6).val);
//
// System.out.println(LCA(t1, t2, t4).val);
// System.out.println(LCA(t1, t2, t6).val);
// System.out.println(LCA(t1, t4, t6).val);
// System.out.println(LCA(t1, t4, t7).val);
// System.out.println(LCA(t1, t3, t6).val);
// System.out.println(LCA(t1, t6, t6).val);
//System.out.println(getMaxDistanceRec(t1));
//System.out.println(isSame(r1, t1));
// System.out.println(isAVLRec(r1));
//
// preorderTraversalRec(r1);
// //mirrorRec(r1);
// //TreeNode r1Mirror = mirror(r1);
//
// TreeNode r1MirrorCopy = mirrorCopy(r1);
// System.out.println();
// //preorderTraversalRec(r1Mirror);
// preorderTraversalRec(r1MirrorCopy);
//
// System.out.println();
//
// System.out.println(isMirrorRec(r1, r1MirrorCopy));
// System.out.println(isMirror(r1, r1MirrorCopy));
//System.out.println(getNodeNumKthLevelRec(r1, 5));
//System.out.println(getNodeNumLeaf(r1));
// System.out.println(getNodeNumRec(null));
// System.out.println(getNodeNum(r1));
//System.out.println(getDepthRec(null));
// System.out.println(getDepth(r1));
//
// preorderTraversalRec(r1);
// System.out.println();
// preorderTraversal(r1);
// System.out.println();
// inorderTraversalRec(r1);
//
// System.out.println();
// inorderTraversal(r1);
// postorderTraversalRec(r1);
// System.out.println();
// postorderTraversal(r1);
// System.out.println();
// levelTraversal(r1);
//
// System.out.println();
// levelTraversalRec(r1);
// TreeNode ret = convertBST2DLLRec(r1);
// while (ret != null) {
// System.out.print(ret.val + " ");
// ret = ret.right;
// }
// TreeNode ret2 = convertBST2DLL(r1);
// while (ret2.right != null) {
// ret2 = ret2.right;
// }
//
// while (ret2 != null) {
// System.out.print(ret2.val + " ");
// ret2 = ret2.left;
// }
//
// TreeNode ret = convertBST2DLL(r1);
// while (ret != null) {
// System.out.print(ret.val + " ");
// ret = ret.right;
// }
// System.out.println();
// System.out.println(findLongest(r1));
// System.out.println();
// System.out.println(findLongest2(r1));
// test the rebuildBinaryTreeRec.
//test_rebuildBinaryTreeRec();
System.out.println(isCompleteBinaryTreeRec(t1));
System.out.println(isCompleteBinaryTree(t1));
}
public static void test_rebuildBinaryTreeRec() {
ArrayList<Integer> list1 = new ArrayList<Integer>();
list1.add(1);
list1.add(2);
list1.add(4);
list1.add(5);
list1.add(3);
list1.add(6);
list1.add(7);
list1.add(8);
ArrayList<Integer> list2 = new ArrayList<Integer>();
list2.add(4);
list2.add(2);
list2.add(5);
list2.add(1);
list2.add(3);
list2.add(7);
list2.add(6);
list2.add(8);
TreeNode root = rebuildBinaryTreeRec(list1, list2);
preorderTraversalRec(root);
System.out.println();
postorderTraversalRec(root);
}
private static class TreeNode{
int val;
TreeNode left;
TreeNode right;
public TreeNode(int val){
this.val = val;
left = null;
right = null;
}
}
/*
* null返回0，然后把左右子树的size加上即可。
* */
public static int getNodeNumRec(TreeNode root) {
if (root == null) {
return 0;
}
return getNodeNumRec(root.left) + getNodeNumRec(root.right) + 1;
}
/**
* 求二叉树中的节点个数迭代解法O(n)：基本思想同LevelOrderTraversal，
* 即用一个Queue，在Java里面可以用LinkedList来模拟
*/
public static int getNodeNum(TreeNode root) {
if (root == null) {
return 0;
}
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
int cnt = 0;
while (!q.isEmpty()) {
TreeNode node = q.poll();
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
cnt++;
}
return cnt;
}
public static int getDepthRec(TreeNode root) {
if (root == null) {
return -1;
}
return Math.max(getDepthRec(root.left), getDepthRec(root.right)) + 1;
}
/*
* 可以用 level LevelOrderTraversal 来实现，我们用一个dummyNode来分隔不同的层，这样即可计算出实际的depth.
* 1
/ \
2 3
/ \ \
4 5 6
*
* 在队列中如此排列： 1, dummy, 2, 3, dummy, 4, 5, 5, dummy
*
*/
public static int getDepth(TreeNode root) {
if (root == null) {
return 0;
}
TreeNode dummy = new TreeNode(0);
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
q.offer(dummy);
int depth = -1;
while (!q.isEmpty()) {
TreeNode curr = q.poll();
if (curr == dummy) {
depth++;
if (!q.isEmpty()) { // 使用DummyNode来区分不同的层， 如果下一层不是为空，则应该在尾部加DummyNode.
q.offer(dummy);
}
}
if (curr.left != null) {
q.offer(curr.left);
}
if (curr.right != null) {
q.offer(curr.right);
}
}
return depth;
}
/*
* 3. 前序遍历，中序遍历，后序遍历: preorderTraversalRec, preorderTraversal, inorderTraversalRec, postorderTraversalRec
* (https://en.wikipedia.org/wiki/Tree_traversal#Pre-order_2)
* */
public static void preorderTraversalRec(TreeNode root) {
if (root == null) {
return;
}
System.out.print(root.val + " ");
preorderTraversalRec(root.left);
preorderTraversalRec(root.right);
}
/*
* 前序遍历，Iteration 算法. 把根节点存在stack中。
* */
public static void preorderTraversal(TreeNode root) {
if (root == null) {
return;
}
Stack<TreeNode> s = new Stack<TreeNode>();
s.push(root);
while (!s.isEmpty()) {
TreeNode node = s.pop();
System.out.print(node.val + " ");
if (node.right != null) { //
s.push(node.right);
}
// 我们需要先压入右节点，再压入左节点，这样就可以先弹出左节点。
if (node.left != null) {
s.push(node.left);
}
}
}
/*
* 中序遍历
* */
public static void inorderTraversalRec(TreeNode root) {
if (root == null) {
return;
}
inorderTraversalRec(root.left);
System.out.print(root.val + " ");
inorderTraversalRec(root.right);
}
/**
* 中序遍历迭代解法 ，用栈先把根节点的所有左孩子都添加到栈内，
* 然后输出栈顶元素，再处理栈顶元素的右子树
* http://www.youtube.com/watch?v=50v1sJkjxoc
*
* 还有一种方法能不用递归和栈，基于线索二叉树的方法，较麻烦以后补上
* http://www.geeksforgeeks.org/inorder-tree-traversal-without-recursion-and-without-stack/
*/
public static void inorderTraversal(TreeNode root) {
if (root == null) {
return;
}
Stack<TreeNode> s = new Stack<TreeNode>();
TreeNode cur = root;
while(true) {
// 把当前节点的左节点都push到栈中.
while (cur != null) {
s.push(cur);
cur = cur.left;
}
if (s.isEmpty()) {
break;
}
// 因为此时已经没有左孩子了，所以输出栈顶元素
cur = s.pop();
System.out.print(cur.val + " ");
// 准备处理右子树
cur = cur.right;
}
}
// 后序遍历
/*
* 1
/ \
2 3
/ \ \
4 5 6
if put into the stack directly, then it should be:
1, 2, 4, 5, 3, 6 in the stack.
when pop, it should be: 6, 3, 5, 4, 2, 1
if I
* */
public static void postorderTraversalRec(TreeNode root) {
if (root == null) {
return;
}
postorderTraversalRec(root.left);
postorderTraversalRec(root.right);
System.out.print(root.val + " ");
}
/**
* 后序遍历迭代解法
* http://www.youtube.com/watch?v=hv-mJUs5mvU
* http://blog.csdn.net/tang_jin2015/article/details/8545457
* 从左到右的后序 与从右到左的前序的逆序是一样的，所以就简单喽！ 哈哈
* 用另外一个栈进行翻转即可喽
*/
public static void postorderTraversal(TreeNode root) {
if (root == null) {
return;
}
Stack<TreeNode> s = new Stack<TreeNode>();
Stack<TreeNode> out = new Stack<TreeNode>();
s.push(root);
while(!s.isEmpty()) {
TreeNode cur = s.pop();
out.push(cur);
if (cur.left != null) {
s.push(cur.left);
}
if (cur.right != null) {
s.push(cur.right);
}
}
while(!out.isEmpty()) {
System.out.print(out.pop().val + " ");
}
}
/*
* 分层遍历二叉树（按层次从上往下，从左往右）迭代
* 其实就是广度优先搜索，使用队列实现。队列初始化，将根节点压入队列。当队列不为空，进行如下操作：弹出一个节点
* ，访问，若左子节点或右子节点不为空，将其压入队列
* */
public static void levelTraversal(TreeNode root) {
if (root == null) {
return;
}
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
while (!q.isEmpty()) {
TreeNode cur = q.poll();
System.out.print(cur.val + " ");
if (cur.left != null) {
q.offer(cur.left);
}
if (cur.right != null) {
q.offer(cur.right);
}
}
}
public static void levelTraversalRec(TreeNode root) {
ArrayList<ArrayList<Integer>> ret = new ArrayList<ArrayList<Integer>>();
levelTraversalVisit(root, 0, ret);
System.out.println(ret);
}
/**
* 分层遍历二叉树（递归）
* 很少有人会用递归去做level traversal
* 基本思想是用一个大的ArrayList，里面包含了每一层的ArrayList。
* 大的ArrayList的size和level有关系
*
* http://discuss.leetcode.com/questions/49/binary-tree-level-order-traversal#answer-container-2543
*/
public static void levelTraversalVisit(TreeNode root, int level, ArrayList<ArrayList<Integer>> ret) {
if (root == null) {
return;
}
// 如果ArrayList的层数不够用， 则新添加一层
// when size = 3, level: 0, 1, 2
if (level >= ret.size()) {
ret.add(new ArrayList<Integer>());
}
// visit 当前节点
ret.get(level).add(root.val);
// 将左子树， 右子树添加到对应的层。
levelTraversalVisit(root.left, level + 1, ret);
levelTraversalVisit(root.right, level + 1, ret);
}
/*
* 题目要求：将二叉查找树转换成排序的双向链表，不能创建新节点，只调整指针。
查找树的结点定义如下：
既然是树，其定义本身就是递归的，自然用递归算法处理就很容易。将根结点的左子树和右子树转换为有序的双向链表，
然后根节点的left指针指向左子树结果的最后一个结点，同时左子树最后一个结点的right指针指向根节点；
根节点的right指针指向右子树结果的第一个结点，
同时右子树第一个结点的left指针指向根节点。
* */
public static TreeNode convertBST2DLLRec(TreeNode root) {
return convertBST2DLLRecHelp(root)[0];
}
/*
* ret[0] 代表左指针
* ret[1] 代表右指针
* */
public static TreeNode[] convertBST2DLLRecHelp(TreeNode root) {
TreeNode[] ret = new TreeNode[2];
ret[0] = null;
ret[1] = null;
if (root == null) {
return ret;
}
if (root.left != null) {
TreeNode left[] = convertBST2DLLRecHelp(root.left);
left[1].right = root; // 将左子树的尾节点连接到根
root.left = left[1];
ret[0] = left[0];
} else {
ret[0] = root; // 左节点返回root.
}
if (root.right != null) {
TreeNode right[] = convertBST2DLLRecHelp(root.right);
right[0].left = root; // 将右子树的头节点连接到根
root.right = right[0];
ret[1] = right[1];
} else {
ret[1] = root; // 右节点返回root.
}
return ret;
}
/**
* 将二叉查找树变为有序的双向链表 迭代解法
* 类似inOrder traversal的做法
*/
public static TreeNode convertBST2DLL(TreeNode root) {
while (root == null) {
return null;
}
TreeNode pre = null;
Stack<TreeNode> s = new Stack<TreeNode>();
TreeNode cur = root;
TreeNode head = null; // 链表头
while (true) {
while (cur != null) {
s.push(cur);
cur = cur.left;
}
// if stack is empty, just break;
if (s.isEmpty()) {
break;
}
cur = s.pop();
if (head == null) {
head = cur;
}
// link pre and cur.
cur.left = pre;
if (pre != null) {
pre.right = cur;
}
// 左节点已经处理完了，处理右节点
cur = cur.right;
pre = cur;
}
return root;
}
/*
* * 6. 求二叉树第K层的节点个数：getNodeNumKthLevelRec, getNodeNumKthLevel
* */
public static int getNodeNumKthLevel(TreeNode root, int k) {
if (root == null || k <= 0) {
return 0;
}
int level = 0;
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
TreeNode dummy = new TreeNode(0);
int cnt = 0; // record the size of the level.
q.offer(dummy);
while (!q.isEmpty()) {
TreeNode node = q.poll();
if (node == dummy) {
level++;
if (level == k) {
return cnt;
}
cnt = 0; // reset the cnt;
if (q.isEmpty()) {
break;
}
q.offer(dummy);
continue;
}
cnt++;
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
return 0;
}
/*
* * 6. 求二叉树第K层的节点个数：getNodeNumKthLevelRec, getNodeNumKthLevel
* */
public static int getNodeNumKthLevelRec(TreeNode root, int k) {
if (root == null || k <= 0) {
return 0;
}
if (k == 1) {
return 1;
}
// 将左子树及右子树在K层的节点个数相加.
return getNodeNumKthLevelRec(root.left, k - 1) + getNodeNumKthLevelRec(root.right, k - 1);
}
/*
* 7. getNodeNumLeafRec 把左子树和右子树的叶子节点加在一起即可
* */
public static int getNodeNumLeafRec(TreeNode root) {
if (root == null) {
return 0;
}
if (root.left == null && root.right == null) {
return 1;
}
return getNodeNumLeafRec(root.left) + getNodeNumLeafRec(root.right);
}
/* 7. getNodeNumLeaf
* 随便使用一种遍历方法都可以，比如，中序遍历。
* inorderTraversal，判断是不是叶子节点。
* */
public static int getNodeNumLeaf(TreeNode root) {
if (root == null) {
return 0;
}
int cnt = 0;
// we can use inorderTraversal travesal to do it.
Stack<TreeNode> s = new Stack<TreeNode>();
TreeNode cur = root;
while (true) {
while (cur != null) {
s.push(cur);
cur = cur.left;
}
if (s.isEmpty()) {
break;
}
// all the left child has been put into the stack, let's deal with the
// current node.
cur = s.pop();
if (cur.left == null && cur.right == null) {
cnt++;
}
cur = cur.right;
}
return cnt;
}
/*
* 8. 判断两棵二叉树是否相同的树。
* 递归解法：
* （1）如果两棵二叉树都为空，返回真
* （2）如果两棵二叉树一棵为空，另一棵不为空，返回假
* （3）如果两棵二叉树都不为空，如果对应的左子树和右子树都同构返回真，其他返回假
* */
public static boolean isSameRec(TreeNode r1, TreeNode r2) {
// both are null.
if (r1 == null && r2 == null) {
return true;
}
// one is null.
if (r1 == null || r2 == null) {
return false;
}
// 1. the value of the root should be the same;
// 2. the left tree should be the same.
// 3. the right tree should be the same.
return r1.val == r2.val &&
isSameRec(r1.left, r2.left) && isSameRec(r1.right, r2.right);
}
/*
* 8. 判断两棵二叉树是否相同的树。
* 迭代解法
* 我们直接用中序遍历来比较就好啦
* */
public static boolean isSame(TreeNode r1, TreeNode r2) {
// both are null.
if (r1 == null && r2 == null) {
return true;
}
// one is null.
if (r1 == null || r2 == null) {
return false;
}
Stack<TreeNode> s1 = new Stack<TreeNode>();
Stack<TreeNode> s2 = new Stack<TreeNode>();
TreeNode cur1 = r1;
TreeNode cur2 = r2;
while (true) {
while (cur1 != null && cur2 != null) {
s1.push(cur1);
s2.push(cur2);
cur1 = cur1.left;
cur2 = cur2.left;
}
if (cur1 != null || cur2 != null) {
return false;
}
if (s1.isEmpty() && s2.isEmpty()) {
break;
}
cur1 = s1.pop();
cur2 = s2.pop();
if (cur1.val != cur2.val) {
return false;
}
cur1 = cur1.right;
cur2 = cur2.right;
}
return true;
}
/*
*
* 9. 判断二叉树是不是平衡二叉树：isAVLRec
* 1. 左子树，右子树的高度差不能超过1
* 2. 左子树，右子树都是平衡二叉树。
*
*/
public static boolean isAVLRec(TreeNode root) {
if (root == null) {
return true;
}
// 左子树，右子树都必须是平衡二叉树。
if (!isAVLRec(root.left) || !isAVLRec(root.right)) {
return false;
}
int dif = Math.abs(getDepthRec(root.left) - getDepthRec(root.right));
if (dif > 1) {
return false;
}
return true;
}
/**
* 10. 求二叉树的镜像 递归解法：
*
* (1) 破坏原来的树
*
* 1 1
* / \
* 2 -----> 2
* \ /
* 3 3
* */
public static TreeNode mirrorRec(TreeNode root) {
if (root == null) {
return null;
}
// 先把左右子树分别镜像,并且交换它们
TreeNode tmp = root.right;
root.right = mirrorRec(root.left);
root.left = mirrorRec(tmp);
return root;
}
/**
* 10. 求二叉树的镜像 Iterator解法：
*
* (1) 破坏原来的树
*
* 1 1
* / \
* 2 -----> 2
* \ /
* 3 3
*
* 应该可以使用任何一种Traversal 方法。
* 我们现在可以试看看使用最简单的前序遍历。
* */
public static TreeNode mirror(TreeNode root) {
if (root == null) {
return null;
}
Stack<TreeNode> s = new Stack<TreeNode>();
s.push(root);
while (!s.isEmpty()) {
TreeNode cur = s.pop();
// 交换当前节点的左右节点
TreeNode tmp = cur.left;
cur.left = cur.right;
cur.right = tmp;
// traversal 左节点，右节点。
if (cur.right != null) {
s.push(cur.right);
}
if (cur.left != null) {
s.push(cur.left);
}
}
return root;
}
/**
* 10. 求二叉树的镜像 Iterator解法：
*
* (2) 创建一个新的树
*
* 1 1
* / \
* 2 -----> 2
* \ /
* 3 3
*
* 应该可以使用任何一种Traversal 方法。
* 我们现在可以试看看使用最简单的前序遍历。
* 前序遍历我们可以立刻把新建好的左右节点创建出来，比较方便
* */
public static TreeNode mirrorCopy(TreeNode root) {
if (root == null) {
return null;
}
Stack<TreeNode> s = new Stack<TreeNode>();
Stack<TreeNode> sCopy = new Stack<TreeNode>();
s.push(root);
TreeNode rootCopy = new TreeNode(root.val);
sCopy.push(rootCopy);
while (!s.isEmpty()) {
TreeNode cur = s.pop();
TreeNode curCopy = sCopy.pop();
// traversal 左节点，右节点。
if (cur.right != null) {
// copy 在这里做比较好，因为我们可以容易地找到它的父节点
TreeNode leftCopy = new TreeNode(cur.right.val);
curCopy.left = leftCopy;
s.push(cur.right);
sCopy.push(curCopy.left);
}
if (cur.left != null) {
// copy 在这里做比较好，因为我们可以容易地找到它的父节点
TreeNode rightCopy = new TreeNode(cur.left.val);
curCopy.right = rightCopy;
s.push(cur.left);
sCopy.push(curCopy.right);
}
}
return rootCopy;
}
/**
* 10. 求二叉树的镜像 递归解法：
*
* (1) 不破坏原来的树，新建一个树
*
* 1 1
* / \
* 2 -----> 2
* \ /
* 3 3
* */
public static TreeNode mirrorCopyRec(TreeNode root) {
if (root == null) {
return null;
}
// 先把左右子树分别镜像,并且把它们连接到新建的root节点。
TreeNode rootCopy = new TreeNode(root.val);
rootCopy.left = mirrorCopyRec(root.right);
rootCopy.right = mirrorCopyRec(root.left);
return rootCopy;
}
/*
* 10.1. 判断两个树是否互相镜像
* (1) 根必须同时为空，或是同时不为空
*
* 如果根不为空:
* (1).根的值一样
* (2).r1的左树是r2的右树的镜像
* (3).r1的右树是r2的左树的镜像
* */
public static boolean isMirrorRec(TreeNode r1, TreeNode r2){
// 如果2个树都是空树
if (r1 == null && r2 == null) {
return true;
}
// 如果其中一个为空，则返回false.
if (r1 == null || r2 == null) {
return false;
}
// If both are not null, they should be:
// 1. have same value for root.
// 2. R1's left tree is the mirror of R2's right tree;
// 3. R2's right tree is the mirror of R1's left tree;
return r1.val == r2.val
&& isMirrorRec(r1.left, r2.right)
&& isMirrorRec(r1.right, r2.left);
}
/*
* 10.1. 判断两个树是否互相镜像 Iterator 做法
* (1) 根必须同时为空，或是同时不为空
*
* 如果根不为空:
* traversal 整个树，判断它们是不是镜像，每次都按照反向来traversal
* (1). 当前节点的值相等
* (2). 当前节点的左右节点要镜像，
* 无论是左节点，还是右节点，对应另外一棵树的镜像位置，可以同时为空，或是同时不为空，但是不可以一个为空，一个不为空。
* */
public static boolean isMirror(TreeNode r1, TreeNode r2){
// 如果2个树都是空树
if (r1 == null && r2 == null) {
return true;
}
// 如果其中一个为空，则返回false.
if (r1 == null || r2 == null) {
return false;
}
Stack<TreeNode> s1 = new Stack<TreeNode>();
Stack<TreeNode> s2 = new Stack<TreeNode>();
s1.push(r1);
s2.push(r2);
while (!s1.isEmpty() && !s2.isEmpty()) {
TreeNode cur1 = s1.pop();
TreeNode cur2 = s2.pop();
// 弹出的节点的值必须相等
if (cur1.val != cur2.val) {
return false;
}
// tree1的左节点，tree2的右节点，可以同时不为空，也可以同时为空，否则返回false.
TreeNode left1 = cur1.left;
TreeNode right1 = cur1.right;
TreeNode left2 = cur2.left;
TreeNode right2 = cur2.right;
if (left1 != null && right2 != null) {
s1.push(left1);
s2.push(right2);
} else if (!(left1 == null && right2 == null)) {
return false;
}
// tree1的左节点，tree2的右节点，可以同时不为空，也可以同时为空，否则返回false.
if (right1 != null && left2 != null) {
s1.push(right1);
s2.push(left2);
} else if (!(right1 == null && left2 == null)) {
return false;
}
}
return true;
}
/*
* 11. 求二叉树中两个节点的最低公共祖先节点：
* Recursion Version:
* LACRec
* 1. If found in the left tree, return the Ancestor.
* 2. If found in the right tree, return the Ancestor.
* 3. If Didn't find any of the node, return null.
* 4. If found both in the left and the right tree, return the root.
* */
public static TreeNode LACRec(TreeNode root, TreeNode node1, TreeNode node2) {
if (root == null || node1 == null || node2 == null) {
return null;
}
// If any of the node is the root, just return the root.
if (root == node1 || root == node2) {
return root;
}
// if no node is in the node, just recursively find it in LEFT and RIGHT tree.
TreeNode left = LACRec(root.left, node1, node2);
TreeNode right = LACRec(root.right, node1, node2);
if (left == null) { // If didn't found in the left tree, then just return it from right.
return right;
} else if (right == null) { // Or if didn't found in the right tree, then just return it from the left side.
return left;
}
// if both right and right found a node, just return the root as the Common Ancestor.
return root;
}
/*
* 11. 求BST中两个节点的最低公共祖先节点：
* Recursive version:
* LCABst
*
* 1. If found in the left tree, return the Ancestor.
* 2. If found in the right tree, return the Ancestor.
* 3. If Didn't find any of the node, return null.
* 4. If found both in the left and the right tree, return the root.
* */
public static TreeNode LCABstRec(TreeNode root, TreeNode node1, TreeNode node2) {
if (root == null || node1 == null || node2 == null) {
return null;
}
// If any of the node is the root, just return the root.
if (root == node1 || root == node2) {
return root;
}
int min = Math.min(node1.val, node2.val);
int max = Math.max(node1.val, node2.val);
// if the values are smaller than the root value, just search them in the left tree.
if (root.val > max) {
return LCABstRec(root.left, node1, node2);
} else if (root.val < min) {
// if the values are larger than the root value, just search them in the right tree.
return LCABstRec(root.right, node1, node2);
}
// if root is in the middle, just return the root.
return root;
}
/*
* 解法1. 记录下path,并且比较之:
* LAC
* http://www.geeksforgeeks.org/lowest-common-ancestor-binary-tree-set-1/
* */
public static TreeNode LCA(TreeNode root, TreeNode r1, TreeNode r2) {
// If the nodes have one in the root, just return the root.
if (root == null || r1 == null || r2 == null) {
return null;
}
ArrayList<TreeNode> list1 = new ArrayList<TreeNode>();
ArrayList<TreeNode> list2 = new ArrayList<TreeNode>();
boolean find1 = LCAPath(root, r1, list1);
boolean find2 = LCAPath(root, r2, list2);
// If didn't find any of the node, just return a null.
if (!find1 || !find2) {
return null;
}
// 注意: 使用Iterator 对于linkedlist可以提高性能。
// 所以 统一使用Iterator 来进行操作。
Iterator<TreeNode> iter1 = list1.iterator();
Iterator<TreeNode> iter2 = list2.iterator();
TreeNode last = null;
while (iter1.hasNext() && iter2.hasNext()) {
TreeNode tmp1 = iter1.next();
TreeNode tmp2 = iter2.next();
if (tmp1 != tmp2) {
return last;
}
last = tmp1;
}
// If never find any node which is different, means Node 1 and Node 2 are the same one.
// so just return the last one.
return last;
}
public static boolean LCAPath(TreeNode root, TreeNode node, ArrayList<TreeNode> path) {
// if didn't find, we should return a empty path.
if (root == null || node == null) {
return false;
}
// First add the root node.
path.add(root);
// if the node is in the left side.
if (root != node
&& !LCAPath(root.left, node, path)
&& !LCAPath(root.right, node, path)
) {
// Didn't find the node. should remove the node added before.
path.remove(root);
return false;
}
// found
return true;
}
/*
* * 12. 求二叉树中节点的最大距离：getMaxDistanceRec
*
* 首先我们来定义这个距离：
* 距离定义为：两个节点间边的数目.
* 如：
* 1
* / \
* 2 3
* \
* 4
* 这里最大距离定义为2，4的距离，为3.
* 求二叉树中节点的最大距离 即二叉树中相距最远的两个节点之间的距离。 (distance / diameter)
* 递归解法：
* 返回值设计：
* 返回1. 深度， 2. 当前树的最长距离
* (1) 计算左子树的深度，右子树深度，左子树独立的链条长度，右子树独立的链条长度
* (2) 最大长度为三者之最：
* a. 通过根节点的链，为左右深度+2
* b. 左子树独立链
* c. 右子树独立链。
*
* (3)递归初始条件：
* 当root == null, depth = -1.maxDistance = -1;
*
*/
public static int getMaxDistanceRec(TreeNode root) {
return getMaxDistanceRecHelp(root).maxDistance;
}
public static Result getMaxDistanceRecHelp(TreeNode root) {
Result ret = new Result(-1, -1);
if (root == null) {
return ret;
}
Result left = getMaxDistanceRecHelp(root.left);
Result right = getMaxDistanceRecHelp(root.right);
// 深度应加1， the depth from the subtree to the root.
ret.depth = Math.max(left.depth, right.depth) + 1;
// 左子树，右子树与根的距离都要加1，所以通过根节点的路径为两边深度+2
int crossLen = left.depth + right.depth + 2;
// 求出cross根的路径，及左右子树的独立路径，这三者路径的最大值。
ret.maxDistance = Math.max(left.maxDistance, right.maxDistance);
ret.maxDistance = Math.max(ret.maxDistance, crossLen);
return ret;
}
private static class Result {
int depth;
int maxDistance;
public Result(int depth, int maxDistance) {
this.depth = depth;
this.maxDistance = maxDistance;
}
}
/*
* 13. 由前序遍历序列和中序遍历序列重建二叉树：rebuildBinaryTreeRec
* We assume that there is no duplicate in the trees.
* For example:
* 1
* / \
* 2 3
* /\ \
* 4 5 6
* /\
* 7 8
*
* PreOrder should be: 1 2 4 5 3 6 7 8
* 根 左子树 右子树
* InOrder should be: 4 2 5 1 3 7 6 8
* 左子树 根 右子树
* */
public static TreeNode rebuildBinaryTreeRec(List<Integer> preOrder, List<Integer> inOrder) {
if (preOrder == null || inOrder == null) {
return null;
}
// If the traversal is empty, just return a NULL.
if (preOrder.size() == 0 || inOrder.size() == 0) {
return null;
}
// we can get the root from the preOrder.
// Because the first one is the root.
// So we just create the root node here.
TreeNode root = new TreeNode(preOrder.get(0));
List<Integer> preOrderLeft;
List<Integer> preOrderRight;
List<Integer> inOrderLeft;
List<Integer> inOrderRight;
// 获得在 inOrder中，根的位置
int rootInIndex = inOrder.indexOf(preOrder.get(0));
preOrderLeft = preOrder.subList(1, rootInIndex + 1);
preOrderRight = preOrder.subList(rootInIndex + 1, preOrder.size());
// 得到inOrder左边的左子树
inOrderLeft = inOrder.subList(0, rootInIndex);
inOrderRight = inOrder.subList(rootInIndex + 1, inOrder.size());
// 通过 Rec 来调用生成左右子树。
root.left = rebuildBinaryTreeRec(preOrderLeft, inOrderLeft);
root.right = rebuildBinaryTreeRec(preOrderRight, inOrderRight);
return root;
}
/*
* 14. 判断二叉树是不是完全二叉树：isCompleteBinaryTree, isCompleteBinaryTreeRec
* 进行level traversal, 一旦遇到一个节点的左节点为空，后面的节点的子节点都必须为空。而且不应该有下一行，其实就是队列中所有的
* 元素都不应该再有子元素。
* */
public static boolean isCompleteBinaryTree(TreeNode root) {
if (root == null) {
return false;
}
TreeNode dummyNode = new TreeNode(0);
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
q.offer(dummyNode);
// if this is true, no node should have any child.
boolean noChild = false;
while (!q.isEmpty()) {
TreeNode cur = q.poll();
if (cur == dummyNode) {
if (!q.isEmpty()) {
q.offer(dummyNode);
}
// Dummy node不需要处理。
continue;
}
if (cur.left != null) {
// 如果标记被设置，则Queue中任何元素不应再有子元素。
if (noChild) {
return false;
}
q.offer(cur.left);
} else {
// 一旦某元素没有左节点或是右节点，则之后所有的元素都不应有子元素。
// 并且该元素不可以有右节点.
noChild = true;
}
if (cur.right != null) {
// 如果标记被设置，则Queue中任何元素不应再有子元素。
if (noChild) {
return false;
}
q.offer(cur.right);
} else {
// 一旦某元素没有左节点或是右节点，则之后所有的元素都不应有子元素。
noChild = true;
}
}
return true;
}
/*
* 14. 判断二叉树是不是完全二叉树：isCompleteBinaryTreeRec
*
*
* 我们可以分解为：
* CompleteBinary Tree 的条件是：
* 1. 左右子树均为Perfect binary tree, 并且两者Height相同
* 2. 左子树为CompleteBinaryTree, 右子树为Perfect binary tree，并且两者Height差1
* 3. 左子树为Perfect Binary Tree,右子树为CompleteBinaryTree, 并且Height 相同
*
* Base 条件：
* (1) root = null: 为perfect & complete BinaryTree, Height -1;
*
* 而 Perfect Binary Tree的条件：
* 左右子树均为Perfect Binary Tree,并且Height 相同。
* */
public static boolean isCompleteBinaryTreeRec(TreeNode root) {
return isCompleteBinaryTreeRecHelp(root).isCompleteBT;
}
private static class ReturnBinaryTree {
boolean isCompleteBT;
boolean isPerfectBT;
int height;
ReturnBinaryTree(boolean isCompleteBT, boolean isPerfectBT, int height) {
this.isCompleteBT = isCompleteBT;
this.isPerfectBT = isPerfectBT;
this.height = height;
}
}
/*
* 我们可以分解为：
* CompleteBinary Tree 的条件是：
* 1. 左右子树均为Perfect binary tree, 并且两者Height相同
* 2. 左子树为CompleteBinaryTree, 右子树为Perfect binary tree，并且两者Height差1
* 3. 左子树为Perfect Binary Tree,右子树为CompleteBinaryTree, 并且Height 相同
*
* Base 条件：
* (1) root = null: 为perfect & complete BinaryTree, Height -1;
*
* 而 Perfect Binary Tree的条件：
* 左右子树均为Perfect Binary Tree,并且Height 相同。
* */
public static ReturnBinaryTree isCompleteBinaryTreeRecHelp(TreeNode root) {
ReturnBinaryTree ret = new ReturnBinaryTree(true, true, -1);
if (root == null) {
return ret;
}
ReturnBinaryTree left = isCompleteBinaryTreeRecHelp(root.left);
ReturnBinaryTree right = isCompleteBinaryTreeRecHelp(root.right);
// 树的高度为左树高度，右树高度的最大值+1
ret.height = 1 + Math.max(left.height, right.height);
// set the isPerfectBT
ret.isPerfectBT = false;
if (left.isPerfectBT && right.isPerfectBT && left.height == right.height) {
ret.isPerfectBT = true;
}
// set the isCompleteBT.
/*
* CompleteBinary Tree 的条件是：
* 1. 左右子树均为Perfect binary tree, 并且两者Height相同(其实就是本树是perfect tree)
* 2. 左子树为CompleteBinaryTree, 右子树为Perfect binary tree，并且两者Height差1
* 3. 左子树为Perfect Binary Tree,右子树为CompleteBinaryTree, 并且Height 相同
* */
ret.isCompleteBT = ret.isPerfectBT
|| (left.isCompleteBT && right.isPerfectBT && left.height == right.height + 1)
|| (left.isPerfectBT && right.isCompleteBT && left.height == right.height);
return ret;
}
/*
* 15. findLongest
* 第一种解法：
* 返回左边最长，右边最长，及左子树最长，右子树最长。
* */
public static int findLongest(TreeNode root) {
if (root == null) {
return -1;
}
TreeNode l = root;
int cntL = 0;
while (l.left != null) {
cntL++;
l = l.left;
}
TreeNode r = root;
int cntR = 0;
while (r.right != null) {
cntR++;
r = r.right;
}
int lmax = findLongest(root.left);
int rmax = findLongest(root.right);
int max = Math.max(lmax, rmax);
max = Math.max(max, cntR);
max = Math.max(max, cntL);
return max;
}
/* 1
* 2 3
* 3 4
* 6 1
* 7
* 9
* 11
* 2
* 14
* */
public static int findLongest2(TreeNode root) {
int [] maxVal = new int[1];
maxVal[0] = -1;
findLongest2Help(root, maxVal);
return maxVal[0];
}
// ret:
// 0: the left side longest,
// 1: the right side longest.
static int maxLen = -1;
static int[] findLongest2Help(TreeNode root, int[] maxVal) {
int[] ret = new int[2];
if (root == null) {
ret[0] = -1;
ret[1] = -1;
return ret;
}
ret[0] = findLongest2Help(root.left, maxVal)[0] + 1;
ret[1] = findLongest2Help(root.right, maxVal)[1] + 1;
//maxLen = Math.max(maxLen, ret[0]);
//maxLen = Math.max(maxLen, ret[1]);
maxVal[0] = Math.max(maxVal[0], ret[0]);
maxVal[0] = Math.max(maxVal[0], ret[1]);
return ret;
}
}
