[LeetCode]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.

思路：交换左右子节点，递归解决。 非递归用栈解决。

小伙伴快来试试看下谷歌是否有戏~~_(:зゝ∠)__(:зゝ∠)__(:зゝ∠)__(:зゝ∠)_

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

非递归

/** * 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)            return nullptr;        stack<TreeNode*> stk;        TreeNode *res = root;        stk.push(root);        while(!stk.empty()){            TreeNode* node = stk.top();            stk.pop();            if(node->left||node->right){ //非叶节点                TreeNode* temp = node->left;                node->left = node->right;                node->right = temp;            }            if(node->left)//注意空指针                stk.push(node->left);            if(node->right)                stk.push(node->right);        }        return res;    }};