算法分析与设计——LeetCode:101. Symmetric Tree

题目

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).For example, this binary tree 

[1,2,2,3,4,4,3]

 is symmetric: 1
/ \
2 2
/ \ / \
3 4 4 3
But the following 

[1,2,2,null,3,null,3]

 is not:
1
/ \
2 2
\ \
3 3
Note:
Bonus points if you could solve it both recursively and iteratively.

/**
 * 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:
    bool isSymmetric(TreeNode* root) {
 
    }
};

思路

方法1用递归逐步判断左右两边是否对称，方法2则利用了队列进行了同样的过程。
代码
方法1/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(
4000
NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isSymmetric(TreeNode* root) {
if (root == NULL) {
return true;
} else {
return isSymmetric(root -> left, root -> right);
}
}

bool isSymmetric(TreeNode* left, TreeNode* right) {
if (left == NULL && right == NULL) {
return true;
} else if (left == NULL || right == NULL) {
return false;
} else if (left -> val != right -> val) {
return false;
}
return isSymmetric(left -> left, right -> right) && isSymmetric(left -> right, right -> left);
}
};方法2

/**
* 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:
bool isSymmetric(TreeNode* root) {
        if (root == NULL) {
            return true;
        }
        queue<TreeNode*> q1, q2;
        q1.push(root -> left);
        q2.push(root -> right);
        while (!q1.empty() && !q2.empty()) {
            TreeNode *left_;
            TreeNode *right_;
            left_ = q1.front();
            right_ = q2.front();
            q1.pop();
            q2.pop();
            if (left_ == NULL && right_ == NULL) {
                continue;
            } else if (left_ == NULL || right_ == NULL) {
                return false;
            } else if (left_ -> val != right_ -> val) {
                return false;
            }
            q1.push(left_ -> right);
            q1.push(left_ -> left);
            q2.push(right_ -> left);
            q2.push(right_ -> right);
        }
        return true;
    }
};