[leetcode] 226. Invert Binary Tree

Invert a binary tree.

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

to

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

Trivia:

This problem was inspired bythis original tweet byMax 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.

这道题是反转二叉树，找出它的镜像，题目难度为easy。

题目本身比较简单，亮点是Trivia的内容。。

可以分别采用深度优先（DFS）和广度优先（BFS）两种策略，所谓深度优先就是采用递归地方法，具体代码：

class Solution {public:    TreeNode* invertTree(TreeNode* root) {        if(!root) return root;        else {            swap(root->left, root->right);            invertTree(root->left);            invertTree(root->right);            return root;        }    }};

广度优先即采用层次序遍历二叉树，这里需要借助队列，具体代码：

class Solution {public:    TreeNode* invertTree(TreeNode* root) {        if(!root) return root;        queue<TreeNode*> level;        level.push(root);        while(!level.empty()) {            int sz = level.size();            for(int i=0; i<sz; i++) {                TreeNode* curNode = level.front();                level.pop();                swap(curNode->left, curNode->right);                if(curNode->left) level.push(curNode->left);                if(curNode->right) level.push(curNode->right);            }        }        return root;    }};