[LeetCode] Symmetric Tree 解题报告

最新推荐文章于 2020-12-28 22:34:43 发布

原创最新推荐文章于 2020-12-28 22:34:43 发布 · 186 阅读

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本文详细介绍了如何通过非递归和递归两种方法判断二叉树是否对称，包括具体实现步骤和代码解析。

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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

» Solve this problem

[解题思路]
非递归解法：按层遍历，每一层检查一下是否对称。

递归解法：

其中左子树和右子树对称的条件：

两个节点值相等，或者都为空
左节点的左子树和右节点的右子树对称
左节点的右子树和右节点的左子树对称

[Code]
非递归解法

1:  bool isSymmetric(TreeNode *root) {  
2:       if(root == NULL) return true;  
3:       vector<TreeNode*> visitQueue;  
4:       visitQueue.push_back(root);  
5:       int curLevel=1;            
6:       while(curLevel >0)  
7:       {  
8:            int i=0;  
9:            while(i<curLevel)  
10:            {  
11:                 TreeNode* p = visitQueue[i];  
12:                 i++;  
13:                 if(p==NULL) continue;  
14:                 visitQueue.push_back(p->left);  
15:                 visitQueue.push_back(p->right);  
16:            }  
17:            int start = 0, end = curLevel-1;  
18:            while(start< end)   
19:            {   
20:                 TreeNode *pl = visitQueue[start];   
21:                 TreeNode *pr = visitQueue[end];   
22:                 int l = pl== NULL? -1:pl->val;   
23:                 int r = pr== NULL? -1: pr->val;   
24:                 if(l!=r)   
25:                      return false;   
26:                 start++;   
27:                 end--;   
28:            }   
29:            visitQueue.erase(visitQueue.begin(), visitQueue.begin() + curLevel);                 
30:            curLevel = visitQueue.size();   
31:       }   
32:       return true;   
33:  }

递归解法

1:    bool isSymmetric(TreeNode *root) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      if(root == NULL) return true;  
5:      return isSym(root->left, root->right);  
6:    }  
7:    bool isSym(TreeNode *left, TreeNode *right)  
8:    {  
9:      if(left == NULL)  
10:        return right ==NULL;  
11:      if(right == NULL)  
12:        return left == NULL;  
13:      if(left->val != right->val)  
14:        return false;  
15:      if(!isSym(left->left, right->right))  
16:        return false;  
17:      if(!isSym(left->right, right->left))  
18:        return false;  
19:      return true;  
20:    }