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;    }};