【二叉树，这题太好玩了】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 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.

解答：

哈哈哈哈哈，太萌了这个题。。。

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

上面的代码还是很有意思的。。。有返回值，但是却没有赋值。。充分利用了递归。

更优雅的解法是这样：

TreeNode* invertTree(TreeNode* root) {    TreeNode* temp;    if(root==NULL) return root;    temp=root->left;    root->left=invertTree(root->right);    root->right=invertTree(temp);    return root;}

另一个人的BFS，也很简单：

TreeNode* invertTree(TreeNode* root) {        if(nullptr == root) return root;        queue<TreeNode*> myQueue;   // our queue to do BFS    myQueue.push(root);         // push very first item - root    while(!myQueue.empty()){    // run until there are nodes in the queue                 TreeNode *node = myQueue.front();  // get element from queue        myQueue.pop();                     // remove element from queue                         if(node->left != nullptr){         // add left kid to the queue if it exists            myQueue.push(node->left);        }                if(node->right != nullptr){        // add right kid             myQueue.push(node->right);        }                // invert left and right pointers              TreeNode* tmp = node->left;        node->left = node->right;        node->right = tmp;    }    return root;}