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;
}
};