leetcode 226. Invert Binary Tree(递归)

Invert a binary tree.

     4
   /   
  2     7
 /    / 
1   3 6   9

to

     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.

题解：很简单的递归。

/**
 * 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==nullptr||(root->left==nullptr&&root->right==nullptr)){
            return root;
        }
        else if(root->left==nullptr){
            root->left=invertTree(root->right);
            root->right=nullptr;
            return root;
        }
        else if(root->right==nullptr){
            root->right=invertTree(root->left);
            root->left=nullptr;
            return root;
        }
        TreeNode* p=root->left;
        root->left=invertTree(root->right);
        root->right=invertTree(p);
        return root;
    }
};