2.27 Invert Binary Tree(Leetcode 226) post-order解法

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public TreeNode invertTree(TreeNode root) {
12         // base case
13         if(root == null){
14             return null;
15         }
16         invertTree(root.left);
17         invertTree(root.right);
18         TreeNode tmp = root.left;
19         root.left = root.right;
20         root.right = tmp;
21         return root;
22     }
23 }时间复杂度 O(n)空间复杂度 O(logn)-O(n)

原文地址：https://www.cnblogs.com/juanqiuxiaoxiao/p/8481746.html