【leetcode】Symmetric Tree-优快云博客

本文链接：https://blog.youkuaiyun.com/fp1355414/article/details/38390599

该博客主要探讨如何检查一个二叉树是否是对称的。通过给出的示例，解释了对称树的特征，并提出了使用递归方法来解决此问题，同时提到了可以使用前序+中序或中序+后序序列化来确定一棵树。

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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.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode *root) {
        
    }
};

对于BST一般是3种遍历与旋转相关的算法题目，仔细观察发现前序+中序遍历然后判断输出是否对称。

相关知识点：前序+中序 or 中序+后序确定一棵树

这里直接使用递归方法来确定

  struct TreeNode {
      int val;
      TreeNode *left;
      TreeNode *right;
      TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  };
 
class Solution {
public:
    bool isSymmetric(TreeNode *root) {
		return root?isSymmetric_2(root->left,root->right):true;
    }
private:
	bool isSymmetric_2(TreeNode *left,TreeNode *right);
};
bool Solution::isSymmetric_2(TreeNode *left,TreeNode *right){
	if (!left&&!right) return true;
	if (!left||!right) return false;
	return left->val==right->val
		&&isSymmetric_2(left->left,right->right)
		&&isSymmetric_2(left->right,right->left);
	
}