Invert Binary Tree

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本文介绍了一种通过递归方式实现的二叉树翻转算法。该算法首先交换根节点的左右子树，然后递归地翻转左子树和右子树，直至整棵树完全翻转。

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题目描述:

Invert a binary tree.

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

     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 fuck off.

思路：

递归反转：
先反转根节点的左子树和右子树 然后再分别反转左子树，右子树各自的下的左子树右子树。依此类推；da

代码实现:

/**
 * 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;
        TreeNode* tmp =root->left;
        root->left = root->right;
        root->right = tmp;
        invertTree(root->left);
        invertTree(root->right);
        return root;
    }
};