LeetCode-101. Symmetric TreeLevel up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.https://leetcode.com/problems/symmetric-tree/

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

Example 1:

Input: root = [1,2,2,3,4,4,3]
Output: true

Example 2:

Input: root = [1,2,2,null,3,null,3]
Output: false

Constraints:

• The number of nodes in the tree is in the range [1, 1000].
• -100 <= Node.val <= 100

Follow up: Could you solve it both recursively and iteratively?

【C++】

/**
 * 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) {}
 * };
 */

递归

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        return root
            ? subtreeSymmetric(root->left,root->right)
            : true;
    }
    bool subtreeSymmetric(TreeNode* root1, TreeNode* root2){
        if (!root1 && !root2) {return true;}
        else if (root1 && root2 && (root1->val == root2->val)){
            return subtreeSymmetric(root1->left,root2->right)
                && subtreeSymmetric(root1->right,root2->left);
        } else {return false;}
    }
};

分步走，更清晰

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        return root
            ? subtreeSymmetric(root->left,root->right)
            : true;
    }
    bool subtreeSymmetric(TreeNode* root1, TreeNode* root2){
        if (!root1 && !root2) {
            return true;
        }
        if (!root1 || !root2) {
            return false;
        }
        if (root1->val != root2->val) {
            return false;
        }
        return subtreeSymmetric(root1->left,root2->right)
            && subtreeSymmetric(root1->right,root2->left);
    }
};

【Java】

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        return root != null
            ? subtreeSymmetric(root.left, root.right)
            : true;
    }
    boolean subtreeSymmetric(TreeNode root1, TreeNode root2){
        if (root1 == null && root2 == null) {return true;}
        else if (root1 != null
                 && root2 != null
                 && (root1.val == root2.val)) {
            return subtreeSymmetric(root1.left,root2.right)
                 && subtreeSymmetric(root1.right,root2.left); 
        } else {return false;}
    }
}

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