这篇文章列出了leetcode中有关二叉树遍历的题目,之前在 二叉树的深搜和广搜中介绍过,这里再重复一下,因为这都是最基本的操作,需要我们熟练掌握。
1,Binary Tree Level Order Traversal
二叉树的广度优先搜索,输出所有节点的值,说的广搜,我们一律用队列来完成,代码如下:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> llist = new ArrayList<List<Integer>>();
List<Integer> list = new ArrayList<Integer>();
Queue<TreeNode> queue = new LinkedList<TreeNode>();
int count = 0;
int helper = 1;
if(root == null) return llist;
queue.offer(root);
while (!queue.isEmpty()) {
TreeNode node = queue.poll();
if(helper > 0){
list.add(node.val);
helper --;
}
if(node.left != null) {
queue.offer(node.left);
count ++;
}
if(node.right != null) {
queue.offer(node.right);
count ++;
}
if(helper == 0) {
llist.add(new ArrayList<Integer>(list));
list.clear();
helper = count;
count = 0;
}
}
return llist;
}
}
2,Binary Tree Level Order Traversal II
给定二叉树:{3,9,20,#,#,15,7},
输出:[[15,7], [9,20], [3]]
这是第一题的变形,仅仅把结果逆序输出,我们用链表中的addFirst()方法就解决了。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
LinkedList<List<Integer>> llist = new LinkedList<List<Integer>>();
List<Integer> list = new ArrayList<Integer>();
Queue<TreeNode> queue = new LinkedList<TreeNode>();
if(root == null) return llist;
queue.offer(root);
int count = 0;
int helper = 1;
while(!queue.isEmpty()) {
TreeNode node = queue.poll();
if(helper > 0) {
list.add(node.val);
helper --;
}
if(node.left != null) {
queue.offer(node.left);
count ++;
}
if(node.right != null) {
queue.offer(node.right);
count ++;
}
if(helper == 0) {
llist.addFirst(new ArrayList<Integer>(list));
list.clear();
helper = count;
count = 0;
}
}
return llist;
}
}
下面是二叉树的前序遍历,中序遍历,后序遍历,都属于深搜,我们用堆栈来完成。
3,Binary Tree Inorder Traversal
二叉树的中序遍历,输出所有节点的值。
代码如下:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<Integer>();
Stack<TreeNode> stack = new Stack<TreeNode>();
if(root == null) return list;
while(root != null || !stack.isEmpty()) {
if(root != null) {
stack.push(root);
root = root.left;
} else {
TreeNode node = stack.pop();
list.add(node.val);
root = node.right;
}
}
return list;
}
}
4,Binary Tree Preorder Traversal
二叉树的前序遍历,输出所有节点的值。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<Integer>();
Stack<TreeNode> stack = new Stack<TreeNode>();
if(root == null) return list;
stack.push(root);
while(!stack.isEmpty()) {
TreeNode tem = stack.pop();
list.add(tem.val);
if(tem.right != null) {
stack.push(tem.right);
}
if(tem.left != null) {
stack.push(tem.left);
}
}
return list;
}
}
5,Binary Tree Postorder Traversal
二叉树的后序遍历,输出所有节点的值。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
LinkedList<Integer> list = new LinkedList<Integer>();
Stack<TreeNode> stack = new Stack<TreeNode>();
if(root == null) return list;
stack.push(root);
while(!stack.isEmpty()) {
TreeNode node = stack.pop();
list.addFirst(node.val);
if(node.left != null) {
stack.push(node.left);
}
if(node.right != null) {
stack.push(node.right);
}
}
return list;
}
}