LeetCode Algorithms #226 <Invert Binary Tree>

invert a binary tree.

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

to

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

思路：

简单的递归调用，不知道Homebrew的作者为什么没做出来。

解：

<span style="font-size:14px;">/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* invertTree(TreeNode* root) {        if(root == NULL)            return NULL;                    invertTree(root->left);        invertTree(root->right);                TreeNode *tempNode = root->left;        root->left = root->right;        root->right = tempNode;                return root;    }};</span>