101. Symmetric Tree

https://leetcode.com/problems/symmetric-tree/

Problem

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1
   / \
  2   2
   \   \
   3    3

Follow up: Solve it both recursively and iteratively.

Solution

1. Recursive

/**
 * 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) {
        if (root == nullptr) return true;
        return isSymmetric(root->left, root->right);
    }
    bool isSymmetric(TreeNode* left, TreeNode* right) {
        if (left == nullptr) return right == nullptr;
        if (right == nullptr) return left == nullptr;
        return (left->val == right->val) && isSymmetric(left->left, right->right) && isSymmetric(left->right, right->left);
    }
};

2. Iterative

/**
 * 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) {
        if (root == nullptr) return true;
        return isSymmetric(root->left, root->right);
    }
    bool isSymmetric(TreeNode* left, TreeNode* right) {
        if (left == nullptr) return right == nullptr;
        if (right == nullptr) return left == nullptr;
        queue<pair<TreeNode*, TreeNode*>> nodes;
        nodes.push({left, right});
        while (!nodes.empty()) {
            auto pair = nodes.front(); nodes.pop();
            if (pair.first == nullptr && pair.second == nullptr) continue;
            if (pair.first == nullptr || pair.second == nullptr) return false;
            if (pair.first->val != pair.second->val) return false;
            nodes.push({pair.first->left, pair.second->right});
            nodes.push({pair.first->right, pair.second->left});
        }
        return true;
    }
};

• #tree

• #dfs

• #bfs