LeetCode 100. Same Tree && 101. Symmetric Tree

1. 题目描述

100
Given two binary trees, write a function to check if they are equal or not.

Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

101
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

1
/ \
2 2
/ \ / \
3 4 4 3
But the following is not:
1
/ \
2 2
\ \
3 3
Note:
Bonus points if you could solve it both recursively and iteratively.

2. 思路

使用递归， 非常简单的思路。

3. code

3.1 same tree

class Solution {
public:
    bool isSameTree(TreeNode *p, TreeNode *q) {
        if ((p == nullptr && q != nullptr) || 
            (p != nullptr && q == nullptr))
            return false;

        if (p == nullptr && q == nullptr)
            return true;

        return (p->val == q->val) && isSameTree(p->left, q->left) && 
                isSameTree(p->right, q->right);
    }
};

3.2 symmetric tree

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if (!root) return true;
        return isEqual(root->left, root->right);
    }

private:
    bool isEqual(TreeNode * p1, TreeNode * p2){
        if (!p1 && p2 || p1 && !p2) return false;
        if (!p1 && !p2) return true;
        return p1->val == p2->val && isEqual(p1->left, p2->right) && isEqual(p1->right, p2->left);
    }
};

4. 大神代码

实际上是一个BFS 搜索的算法， 使用BFS 对左右两个子树进行搜索， 不过一个是从左向右， 另一个是从右向左， nice code

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

        queue<TreeNode*> q1, q2;
        q1.push(root->left);
        q2.push(root->right);
        while (!q1.empty() && !q2.empty()){
            left = q1.front();
            q1.pop();
            right = q2.front();
            q2.pop();
            if (NULL == left && NULL == right)
                continue;
            if (NULL == left || NULL == right)
                return false;
            if (left->val != right->val)
                return false;
            q1.push(left->left);
            q1.push(left->right);
            q2.push(right->right);
            q2.push(right->left);
        }
        return true;
    }
};