LeetCode 101 Symmetric Tree DFS

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

Solution

判断树是否为对称的。我们可以直接对左右子树进行 \(DFS\) 递归

点击查看代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
private:
    bool check(TreeNode* LeftRoot, TreeNode* RightRoot){
        if(!LeftRoot && !RightRoot) return true;
        else if(!LeftRoot || !RightRoot) return false;
        
        if(LeftRoot->val != RightRoot->val) return false;
        
        return check(LeftRoot->right, RightRoot->left) && check(LeftRoot->left, RightRoot->right);
    }
    
public:
    bool isSymmetric(TreeNode* root) {
        if(!root) return true;
        return check(root->left, root->right);
    }
};