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101.对称二叉树.cpp
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114 lines (102 loc) · 2.77 KB
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/*
* @lc app=leetcode.cn id=101 lang=cpp
*
* [101] 对称二叉树
*/
// @lc code=start
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
// 递归实现镜像比较,时间复杂度还行,空间复杂度较高。
bool recursion(TreeNode* left, TreeNode* right){
if (left == nullptr && right == nullptr)
{
return true;
}
else if(left == nullptr)
{
return false;
}
else if(right == nullptr)
{
return false;
}
else
{
if (right->val == left->val && recursion(left->right, right->left) && recursion(left->left, right->right))
{
return true;
}
else
{
return false;
}
}
}
bool isSymmetric(TreeNode* root){
if (root == nullptr)
{
return true;
}
else
{
return recursion(root->left, root->right);
}
}
/*
// 层次遍历的实现,时间和空间复杂度较高。
bool isSymmetric(TreeNode* root) {
queue<TreeNode*> que; // 队列。
que.push(root);
int sum = que.size(); // 记录每一层的指针个数以便从队列取出。
bool sign = true;
while (sign)
{
sign = false;
vector<int> buffer;
for (int i = 0; i < sum; i++)
{
if (que.front())
{
if (sum == 1 || i < sum / 2)
{
buffer.push_back(que.front()->val);
}
else if (buffer.at(sum - 1 - i) != que.front()->val)
{
return false;
}
que.push(que.front()->left);
que.push(que.front()->right);
sign = true;
}
else
{
if (sum == 1 || i < sum / 2)
{
buffer.push_back(INT_MAX);
}
else if (buffer.at(sum - 1 - i) != INT_MAX)
{
return false;
}
que.push(nullptr);
que.push(nullptr);
}
que.pop();
}
sum = que.size();
}
return true;
}
*/
};
// @lc code=end