Classic 150 Interview Questions P101 Symmetric Binary Tree

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Timeline

Timeline

2025-10-23

init

Binary Tree

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Problem:

P101 Symmetric Binary Tree

https://leetcode.cn/problems/symmetric-tree/description/?envType=study-plan-v2&envId=top-interview-150

Recursion, symmetric comparison

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

Non-recursive approach, iterative, enqueue nodes twice

#include <queue>
using std::queue;

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
	TreeNode *left, *right;

        if (root == nullptr) return true;

        queue<TreeNode*> que;
        que.push(root->left);
        que.push(root->right);


        while (!que.empty()) {
            left = que.front();
            que.pop();
            right = que.front();
            que.pop();

            if (left == nullptr && right == nullptr) continue;
            if (left == nullptr || right == nullptr) return false;
            if (left->val != right->val) return false;

            // Note the symmetric enqueue order here
            que.push(left->left);
            que.push(right->right);
            que.push(left->right);
            que.push(right->left);
        }
        return true;
    }
};

leetcode hot 100 rewrite

class Solution {
    private:
        bool isSame(TreeNode *p, TreeNode *q)
        {
                if ((p && !q) || (!p && q))
                        return false;
                if (!p && !q)
                        return true;

                if (p->val == q->val)
                        return isSame(p->left, q->right) && isSame(p->right, q->left);
                else
                        return false;
        }

    public:
        bool isSymmetric(TreeNode *root)
        {
                // root is not null

                return isSame(root->left, root->right);
        }
};