101. Symmetric Tree

Recursion

/*
 * @lc app=leetcode id=101 lang=cpp
 *
 * [101] Symmetric Tree
 *
 * https://leetcode.com/problems/symmetric-tree/description/
 *
 * algorithms
 * Easy (57.89%)
 * Likes:    15947
 * Dislikes: 410
 * Total Accepted:    2.4M
 * Total Submissions: 4M
 * Testcase Example:  '[1,2,2,3,4,4,3]'
 *
 * 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?
 */

// @lc code=start
/**
 * 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 {
public:
    bool isSymmetric(TreeNode* root) {
        return isSame(root, root);
    }
    bool isSame(TreeNode* t1, TreeNode* t2) {
        if (!t1 && !t2) return true;
        if (!t1 || !t2) return false;
        if (t1->val != t2->val) return false;
        return isSame(t1->left, t2->right) && isSame(t1->right, t2->left);
    }
};
// @lc code=end

• T: $O(N)$
• S: $O(N)$

iteration

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        queue<TreeNode*> q;
        q.push(root);
        q.push(root);
        while (!q.empty()) {
            TreeNode* t1 = q.front(); q.pop();
            TreeNode* t2 = q.front(); q.pop();
            if (!t1 && !t2) continue;
            if (!t1 || !t2) return false;
            if (t1->val != t2->val) return false;
            q.push(t1->left);
            q.push(t2->right);
            q.push(t1->right);
            q.push(t2->left);
        }
        return true;
    }
};

• T: $O(N)$
• S: $O(N)$