标签：binary 左右 cpp null root pac rtt 左右子树 article
该题比較简单。递归交换每个节点的左右子树就可以。
class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
if(root == NULL)
return NULL;
TreeNode* tmp = root -> left;
root -> left = invertTree(root -> right);
root -> right = invertTree(tmp);
}
};标签：binary 左右 cpp null root pac rtt 左右子树 article
原文地址：http://www.cnblogs.com/jhcelue/p/6815934.html
