leetcode 226. Invert Binary Tree

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

4
/   \
7     2
/ \   / \
9   6 3   1
二叉树的递归实现：
[code]<span style="font-size:18px;">/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     struct TreeNode *left;
*     struct TreeNode *right;
* };
*/
struct TreeNode* invertTree(struct TreeNode* root) {
if(root==NULL) return root;
invertTree(root->left);
invertTree(root->right);
struct TreeNode* T;
T=root->left;
root->left=root->right;
root->right=T;
return root;
}</span>
<span style="font-size:18px;">然后是非递归形式的，使用队列实现,：
</span><pre name="code" class="cpp"><span style="font-size:18px;">/**
* 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) {
queue<TreeNode*> q;
if(root) q.push(root);
struct TreeNode *c,*tmp;
while(!q.empty())
{
c=q.front();
q.pop();
tmp=c->left;
c->left=c->right;
c->right=tmp;
if(c->left) q.push(c->left);
if(c->right) q.push(c->right);
}
return root;
}
};</span>