[leetCode] Symmetric Tree

最新推荐文章于 2019-05-29 13:59:21 发布

转载最新推荐文章于 2019-05-29 13:59:21 发布 · 53 阅读

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原文链接：http://www.cnblogs.com/vincently/p/4232530.html

本文介绍了一种检查二叉树是否为中心对称的方法，提供了两种实现方式：递归和迭代。递归方法通过比较左右子树来判断对称性，而迭代方法则采用层次遍历的方式进行检查。

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

For example, this binary tree is symmetric:

But the following is not:

Note:
Bonus points if you could solve it both recursively and iteratively.

思路一：递归思想。时间复杂度O(n)，空间复杂度O(logN)

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool isSymmetric(TreeNode *root) {
13         if (root == NULL) return true;
14         return isSymmetric(root->left, root->right);
15     }
16     bool isSymmetric(TreeNode *root1, TreeNode *root2) {
17         if (!root1 && !root2) return true;
18         if (!root1 || !root2) return false;
19         if (root1->val != root2->val) return false;
20         
21         return isSymmetric(root1->left, root2->right) && isSymmetric(root1->right, root2->left);
22     }
23 };

思路二：迭代。层次遍历的思想。时间复杂度O(n),空间复杂度O(logN）

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool isSymmetric(TreeNode *root) {
13         if (root == NULL) return true;
14         queue<TreeNode *> q;
15         q.push(root->left);
16         q.push(root->right);
17         
18         while (!q.empty()) {
19             TreeNode *p1 = q.front();
20             q.pop();
21             TreeNode *p2 = q.front();
22             q.pop();
23             
24             if (!p1 && !p2) continue;
25             if (!p1 || !p2) return false;
26             if (p1->val != p2->val) return false;
27             
28             q.push(p1->left);
29             q.push(p2->right);
30             
31             q.push(p1->right);
32             q.push(p2->left);
33         }
34         
35         return true;
36     }
37  
38 };

转载于:https://www.cnblogs.com/vincently/p/4232530.html