java数据结构与算法4---二叉树
java数据结构与算法4---二叉树
二叉树介绍
二叉树：每个节点只含有左右两个分支的叫做二叉树。
二叉树练习
题目一：实现二叉树的先序、中序、后序遍历，包括递归方式和非递归方式
思路：先序--->打印、左递归、右递归；中序--->左递归、打印、右递归；后序--->左递归、右递归、打印。
代码实现如下：
//二叉树先序、中序、后序遍历算法（递归&非递归）
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int data) {
this.value = data;
}
}
public static void preOrderRecur(Node head) {
if (head == null) {
return;
}
System.out.print(head.value + " ");
preOrderRecur(head.left);
preOrderRecur(head.right);
}
public static void inOrderRecur(Node head) {
if (head == null) {
return;
}
inOrderRecur(head.left);
System.out.print(head.value + " ");
inOrderRecur(head.right);
}
public static void posOrderRecur(Node head) {
if (head == null) {
return;
}
posOrderRecur(head.left);
posOrderRecur(head.right);
System.out.print(head.value + " ");
}
public static void preOrderUnRecur(Node head) {
System.out.print("pre-order: ");
if (head != null) {
Stack stack = new Stack();
stack.add(head);
while (!stack.isEmpty()) {
head = stack.pop();
System.out.print(head.value + " ");
if (head.right != null) {
stack.push(head.right);
}
if (head.left != null) {
stack.push(head.left);
}
}
}
System.out.println();
}
public static void inOrderUnRecur(Node head) {
System.out.print("in-order: ");
if (head != null) {
Stack stack = new Stack();
while (!stack.isEmpty() || head != null) {
if (head != null) {
stack.push(head);
head = head.left;
} else {
head = stack.pop();
System.out.print(head.value + " ");
head = head.right;
}
}
}
System.out.println();
}
public static void posOrderUnRecur1(Node head) {
System.out.print("pos-order: ");
if (head != null) {
Stack s1 = new Stack();
Stack s2 = new Stack();
s1.push(head);
while (!s1.isEmpty()) {
head = s1.pop();
s2.push(head);
if (head.left != null) {
s1.push(head.left);
}
if (head.right != null) {
s1.push(head.right);
}
}
while (!s2.isEmpty()) {
System.out.print(s2.pop().value + " ");
}
}
System.out.println();
}
public static void posOrderUnRecur2(Node h) {
System.out.print("pos-order: ");
if (h != null) {
Stack stack = new Stack();
stack.push(h);
Node c = null;
while (!stack.isEmpty()) {
c = stack.peek();
if (c.left != null && h != c.left && h != c.right) {
stack.push(c.left);
} else if (c.right != null && h != c.right) {
stack.push(c.right);
} else {
System.out.print(stack.pop().value + " ");
h = c;
}
}
}
System.out.println();
} 题目二：如何直观的打印一颗二叉树
思路：按照自定义规则打印二叉树
代码实现如下：
//打印二叉树
package com.sc.d04;
public class Code_02_PrintBinaryTree {
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int data) {
this.value = data;
}
}
public static void printTree(Node head) {
System.out.println("Binary Tree:");
printInOrder(head, 0, "H", 17);
System.out.println();
}
public static void printInOrder(Node head, int height, String to, int len) {
if (head == null) {
return;
}
printInOrder(head.right, height + 1, "v", len);
String val = to + head.value + to;
int lenM = val.length();
int lenL = (len - lenM) / 2;
int lenR = len - lenM - lenL;
val = getSpace(lenL) + val + getSpace(lenR);
System.out.println(getSpace(height * len) + val);
printInOrder(head.left, height + 1, "^", len);
}
public static String getSpace(int num) {
String space = " ";
StringBuffer buf = new StringBuffer("");
for (int i = 0; i < num; i++) {
buf.append(space);
}
return buf.toString();
}
public static void main(String[] args) {
Node head = new Node(1);
head.left = new Node(-222222222);
head.right = new Node(3);
head.left.left = new Node(Integer.MIN_VALUE);
head.right.left = new Node(55555555);
head.right.right = new Node(66);
head.left.left.right = new Node(777);
printTree(head);
head = new Node(1);
head.left = new Node(2);
head.right = new Node(3);
head.left.left = new Node(4);
head.right.left = new Node(5);
head.right.right = new Node(6);
head.left.left.right = new Node(7);
printTree(head);
head = new Node(1);
head.left = new Node(1);
head.right = new Node(1);
head.left.left = new Node(1);
head.right.left = new Node(1);
head.right.right = new Node(1);
head.left.left.right = new Node(1);
printTree(head);
}
}
题目三：在二叉树中找到一个节点的后继节点
【题目】 现在有一种新的二叉树节点类型如下：
public class Node {public int value;
public Node left;
public Node right;public Node parent;
public Node(int data) { this.value data; }
}该结构比普通二叉树节点结构多了一个指向父节点的parent指针。假设有一 棵Node类型的节点组成的二叉树，树中每个节点的parent指针都正确地指向 自己的父节点，头节点的parent指向null。只给一个在二叉树中的某个节点 node，请实现返回node的后继节点的函数。在二叉树的中序遍历的序列中， node的下一个节点叫作node的后继节点。
思路：求按中序遍历下node的后继节点，需要分情况来讨论。1、当node有right子树时，取right子树的最左端节点为node的后继节点；2、当node无right子树时，向上找node的parent节点，如果node为parent节点的left节点，则parent为node的后继节点；若node不是parent的left节点，则继续往上找使node=parent，parent=parent.parent，至找到满足左节点为止，否则node的后继节点为null。
代码实现如下：
//中遍历下查找后继节点算法
public static class Node {
public int value;
public Node left;
public Node right;
public Node parent;
public Node(int data) {
this.value = data;
}
}
public static Node getSuccessorNode(Node node) {
if (node == null) {
return node;
}
if (node.right != null) {
return getLeftMost(node.right);
} else {
Node parent = node.parent;
while (parent != null && parent.left != node) {
node = parent;
parent = node.parent;
}
return parent;
}
}
public static Node getLeftMost(Node node) {
if (node == null) {
return node;
}
while (node.left != null) {
node = node.left;
}
return node;
}题目四：介绍二叉树的序列化和反序列化
思路：二叉树的序列化---将二叉树的节点值按照规定的遍历方法遍历，最终以String的形式输出没有左子树或右子树的地方用#替换，这样就可以得到二叉树的序列化，这种序列化的特点是将二叉树的特殊结构依旧保留在序列中，但是无法直观的来观察出来；二叉树的反序列化---将序列化的二叉树String，利用序列化时的遍历方法进行反序列化，最终形成二叉树结构。
代码实现如下：
//二叉树的序列化&反序列化算法
package com.sc.d04;
import java.util.LinkedList;
import java.util.Queue;
public class Code_04_SerializeAndReconstructTree {
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int data) {
this.value = data;
}
}
public static String serialByPre(Node head) {
if (head == null) {
return "#!";
}
String res = head.value + "!";
res += serialByPre(head.left);
res += serialByPre(head.right);
return res;
}
public static Node reconByPreString(String preStr) {
String[] values = preStr.split("!");
Queue queue = new LinkedList();
for (int i = 0; i != values.length; i++) {
queue.offer(values[i]);
}
return reconPreOrder(queue);
}
public static Node reconPreOrder(Queue queue) {
String value = queue.poll();
if (value.equals("#")) {
return null;
}
Node head = new Node(Integer.valueOf(value));
head.left = reconPreOrder(queue);
head.right = reconPreOrder(queue);
return head;
}
public static String serialByLevel(Node head) {
if (head == null) {
return "#!";
}
String res = head.value + "!";
Queue queue = new LinkedList();
queue.offer(head);
while (!queue.isEmpty()) {
head = queue.poll();
if (head.left != null) {
res += head.left.value + "!";
queue.offer(head.left);
} else {
res += "#!";
}
if (head.right != null) {
res += head.right.value + "!";
queue.offer(head.right);
} else {
res += "#!";
}
}
return res;
}
public static Node reconByLevelString(String levelStr) {
String[] values = levelStr.split("!");
int index = 0;
Node head = generateNodeByString(values[index++]);
Queue queue = new LinkedList();
if (head != null) {
queue.offer(head);
}
Node node = null;
while (!queue.isEmpty()) {
node = queue.poll();
node.left = generateNodeByString(values[index++]);
node.right = generateNodeByString(values[index++]);
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
return head;
}
public static Node generateNodeByString(String val) {
if (val.equals("#")) {
return null;
}
return new Node(Integer.valueOf(val));
}
// for test -- print tree
public static void printTree(Node head) {
System.out.println("Binary Tree:");
printInOrder(head, 0, "H", 17);
System.out.println();
}
public static void printInOrder(Node head, int height, String to, int len) {
if (head == null) {
return;
}
printInOrder(head.right, height + 1, "v", len);
String val = to + head.value + to;
int lenM = val.length();
int lenL = (len - lenM) / 2;
int lenR = len - lenM - lenL;
val = getSpace(lenL) + val + getSpace(lenR);
System.out.println(getSpace(height * len) + val);
printInOrder(head.left, height + 1, "^", len);
}
public static String getSpace(int num) {
String space = " ";
StringBuffer buf = new StringBuffer("");
for (int i = 0; i < num; i++) {
buf.append(space);
}
return buf.toString();
}
public static void main(String[] args) {
Node head = null;
printTree(head);
String pre = serialByPre(head);
System.out.println("serialize tree by pre-order: " + pre);
head = reconByPreString(pre);
System.out.print("reconstruct tree by pre-order, ");
printTree(head);
String level = serialByLevel(head);
System.out.println("serialize tree by level: " + level);
head = reconByLevelString(level);
System.out.print("reconstruct tree by level, ");
printTree(head);
System.out.println("====================================");
head = new Node(1);
printTree(head);
pre = serialByPre(head);
System.out.println("serialize tree by pre-order: " + pre);
head = reconByPreString(pre);
System.out.print("reconstruct tree by pre-order, ");
printTree(head);
level = serialByLevel(head);
System.out.println("serialize tree by level: " + level);
head = reconByLevelString(level);
System.out.print("reconstruct tree by level, ");
printTree(head);
System.out.println("====================================");
head = new Node(1);
head.left = new Node(2);
head.right = new Node(3);
head.left.left = new Node(4);
head.right.right = new Node(5);
printTree(head);
pre = serialByPre(head);
System.out.println("serialize tree by pre-order: " + pre);
head = reconByPreString(pre);
System.out.print("reconstruct tree by pre-order, ");
printTree(head);
level = serialByLevel(head);
System.out.println("serialize tree by level: " + level);
head = reconByLevelString(level);
System.out.print("reconstruct tree by level, ");
printTree(head);
System.out.println("====================================");
head = new Node(100);
head.left = new Node(21);
head.left.left = new Node(37);
head.right = new Node(-42);
head.right.left = new Node(0);
head.right.right = new Node(666);
printTree(head);
pre = serialByPre(head);
System.out.println("serialize tree by pre-order: " + pre);
head = reconByPreString(pre);
System.out.print("reconstruct tree by pre-order, ");
printTree(head);
level = serialByLevel(head);
System.out.println("serialize tree by level: " + level);
head = reconByLevelString(level);
System.out.print("reconstruct tree by level, ");
printTree(head);
System.out.println("====================================");
}
}
题目五：折纸问题
【题目】 请把一段纸条竖着放在桌子上，然后从纸条的下边向上方对折1次，压出折痕后展开。此时 折痕是凹下去的，即折痕突起的方向指向纸条的背面。如果从纸条的下边向上方连续对折2 次，压出折痕后展开，此时有三条折痕，从上到下依次是下折痕、下折痕和上折痕。给定一 个输入参数N，代表纸条都从下边向上方连续对折N次，请从上到下打印所有折痕的方向。 例如：N 1时，打印： down ；N 2时，打印： down down up。
思路：通过折痕朝向总结我们发现，N相当于二叉树的高，down为二叉树所有的左节点（包含头节点），up为二叉树所有的右节点。从上到下依次打印折痕就是采用中序遍历的方法来打印二叉树。
代码实现如下：
/*
*折纸折痕问题
*转换为二叉树中序遍历打印问题
*/
public static void printAllFolds(int N) {
printProcess(1, N, true);
}
//二叉树中序遍历
public static void printProcess(int i, int N, boolean down) {
if (i > N) {
return;
}
printProcess(i + 1, N, true);
System.out.println(down ? "down " : "up ");
printProcess(i + 1, N, false);
}题目六：判断一棵二叉树是否是平衡二叉树
介绍平衡二叉树：二叉树的左子树与右子树的高度差的绝对值<1
思路：利用递归的思想，求出左子树height与右子树的height，然后左height—右height<1则为平衡二叉树。
代码实现如下：
//判断平衡二叉树算法
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int data) {
this.value = data;
}
}
public static boolean isBalance(Node head) {
boolean[] res = new boolean[1];
res[0] = true;
System.out.println(getHeight(head, 1, res));
return res[0];
}
//利用递归求左右子树height
public static int getHeight(Node head, int level, boolean[] res) {
if (head == null) {
return level;
}
int lH = getHeight(head.left, level + 1, res);
if (!res[0]) {
return level;
}
int rH = getHeight(head.right, level + 1, res);
if (!res[0]) {
return level;
}
if (Math.abs(lH - rH) > 1) {
res[0] = false;
}
return Math.max(lH, rH);
}题目七：判断一棵树是否是二叉搜索树、判断一棵树是否是完全二叉树。
二叉查找树（英语：Binary Search Tree），也称为二叉搜索树、有序二叉树（ordered binary tree）或排序二叉树（sorted binary tree），是指一棵空树或者具有下列性质的二叉树：
- 若任意节点的左子树不空，则左子树上所有节点的值均小于它的根节点的值；
- 若任意节点的右子树不空，则右子树上所有节点的值均大于它的根节点的值；
- 任意节点的左、右子树也分别为二叉查找树；
- 没有键值相等的节点
思路：判断二叉搜索树可以理解为通过中序遍历二叉树得到的遍历值是按递增的顺序排列的；判断完全二叉树需要分情况讨论，1、左子树为null，右子树存在----一定不是完全二叉树 。2、左子树为null且右子树为null || 左子树存在且右子树为null----之后的全为leaf节点。
代码实现如下：
/*
*判断二叉搜索树&判断完全二叉树
*/
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int data) {
this.value = data;
}
}
//中序遍历方法判断二叉搜索树
static int pre=0;
public static boolean myIsBST(Node head)
{
if(head==null)
return false;
myIsBST(head.left);
if(pre cur1.value) {
res = false;
}
pre = cur1;
cur1 = cur1.right;
}
return res;
}
//判断完全二叉树
//采用level遍历方法来判断的
public static boolean isCBT(Node head) {
if (head == null) {
return true;
}
Queue queue = new LinkedList();
boolean leaf = false;
Node l = null;
Node r = null;
queue.offer(head);
while (!queue.isEmpty()) {
head = queue.poll();
l = head.left;
r = head.right;
if ((leaf && (l != null || r != null)) || (l == null && r != null)) {
return false;
}
if (l != null) {
queue.offer(l);
}
if (r != null) {
queue.offer(r);
} else {
leaf = true;
}
}
return true;
}
题目八：已知一棵完全二叉树，求其节点的个数。要求：时间复杂度低于O(N)，N为这棵树的节点个数。
思路：根据完全二叉树的特点，先找到二叉树的最大行数，然后判断根节点的左右子树的maxHeight是否一样，若一样说明左子树是完全maxHeight-1行二叉树；若不一样说明右子树是完全maxHeight-2行二叉树。最后就可以根据完全性来局部求出节点数，相加即为总节点数。
代码实现如下：
//求完全二叉树的节点个数
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int data) {
this.value = data;
}
}
public static int nodeNum(Node head) {
if (head == null) {
return 0;
}
return bs(head, 1, mostLeftLevel(head, 1));
}
public static int bs(Node node, int l, int h) {
if (l == h) {
return 1;
}
if (mostLeftLevel(node.right, l + 1) == h) {
return (1 << (h - l)) + bs(node.right, l + 1, h);
} else {
return (1 << (h - l - 1)) + bs(node.left, l + 1, h);
}
}
public static int mostLeftLevel(Node node, int level) {
while (node != null) {
level++;
node = node.left;
}
return level - 1;
}我将在java数据结构与算法5---前缀树&贪心策略中介绍与前缀树&贪心策略。
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