Invert a binary tree.

Example:

Input:

     4
   /   \
  2     7
 / \   / \
1   3 6   9

Output:

     4
   /   \
  7     2
 / \   / \
9   6 3   1

Trivia:
This problem was inspired by this original tweet by Max Howell:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so f*** off.

翻转二叉树。也是一道不会做，会写homebrew也枉然的题。题干即是题意。例子如下，即层层遍历，把左右子树交换。

两种思路，分别是层序遍历（BFS）和深度遍历（DFS）。

DFS

时间O(n)

空间O(n)

JavaScript实现

 1 /**
 2  * @param {TreeNode} root
 3  * @return {TreeNode}
 4  */
 5 var invertTree = function(root) {
 6     if (root === null) return root;
 7     let left = invertTree(root.left);
 8     let right = invertTree(root.right);
 9     root.left = right;
10     root.right = left;
11     return root;
12 };

 

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     public TreeNode invertTree(TreeNode root) {
18         if (root == null) return root;
19         TreeNode left = invertTree(root.left);
20         TreeNode right = invertTree(root.right);
21         root.left = right;
22         root.right = left;
23         return root;
24     }
25 }

 

BFS

时间O(n)

空间O(n)

JavaScript实现

 1 /**
 2  * @param {TreeNode} root
 3  * @return {TreeNode}
 4  */
 5 var invertTree = function(root) {
 6     if (!root) return root;
 7     let queue = [root];
 8     while (queue.length) {
 9         let current = queue.shift();
10         if (current === null) continue;
11         swap(current);
12         queue.push(current.left);
13         queue.push(current.right);
14     }
15     return root;
16 };
17 
18 var swap = tree => {
19     let temp = tree.left;
20     tree.left = tree.right;
21     tree.right = temp;
22     return tree;
23 };

 

Java实现

 1 class Solution {
 2     public TreeNode invertTree(TreeNode root) {
 3         // corner case
 4         if (root == null) {
 5             return root;
 6         }
 7 
 8         // normal case
 9         Queue<TreeNode> queue = new LinkedList<>();
10         queue.offer(root);
11         while (!queue.isEmpty()) {
12             TreeNode cur = queue.poll();
13             TreeNode temp = cur.left;
14             cur.left = cur.right;
15             cur.right = temp;
16             if (cur.left != null) {
17                 queue.offer(cur.left);
18             }
19             if (cur.right != null) {
20                 queue.offer(cur.right);
21             }
22         }
23         return root;
24     }
25 }

 

LeetCode 题目总结