LeetCode: 226 Invert Binary Tree(easy)

题目：

Invert a binary tree.

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

to

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

代码：

自己的（递归实现）：

/**
 * 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* tem;
        if (!root)
            return 0;             
        tem = root->left;
        root->left = root->right;
        root->right = tem;
        if(root->left)
            invertTree(root->left);
        if(root->right)
            invertTree(root->right);
        return root;
    }
};

别人的：

/**
 * 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){
            swap(root->left, root->right);
            invertTree(root->left);
            invertTree(root->right);     
        }
        return root;
    }
};

自己的（使用队列，非递归实现）:

/**
 * 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*> queue;
        if(!root)
            return 0;
        queue.push(root);
        TreeNode *curNode;
        while(!queue.empty()){
            curNode = queue.front();
            queue.pop();
            swap(curNode->left, curNode->right);
            if(curNode->left)
                queue.push(curNode->left);
            if(curNode->right)
                queue.push(curNode->right);        
        }
        return root;
    }
};

非递归和递归的在LeetCode上运行效率差不多。。。