226. Invert Binary Tree

nvert 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.

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C++递归代码：

/**
* 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 *treeleft=invertTree(root->right);
TreeNode *treeright=invertTree(root->left);
root->left=treeleft;
root->right=treeright;
return root;

}
};

C++非递归代码，主要是BFS宽度优先搜索的思想，利用队列来实现：

class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
if(root==NULL) return NULL;
queue<TreeNode*>tree_que;
tree_que.push(root);
while(tree_que.size()>0)
{
TreeNode *first=tree_que.front();
tree_que.pop();
TreeNode *tmp=first->left;
first->left=first->right;
first->right=tmp;
if(first->left)
tree_que.push(first->left);
if(first->right)
tree_que.push(first->right);
}
return root;
}
};

C#版本,顺便改了下递归顺序，另外C#版本的代码也可以在JAVA环境中运行，主要是NULL大小写的区别，C#和java是没有定义大写的NULL的。

public 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;

}
}