对称二叉树
给你一个二叉树的根节点 root , 检查它是否轴对称。
示例 1:
输入:root = [1,2,2,3,4,4,3]
输出:true
示例 2:
输入:root = [1,2,2,null,3,null,3]
输出:false
提示:
树中节点数目在范围 [1, 1000] 内
-100 <= Node.val <= 100
/**
* Definition for a binary tree node.
* 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 isSymmetric(TreeNode* root) {
std::queue<TreeNode*> q;
if(root == nullptr){
return true;
}
q.push(root->left);
q.push(root->right);
while(!q.empty()){
auto left = q.front();
q.pop();
auto right = q.front();
q.pop();
if(left == nullptr && right == nullptr){
continue;
}
if(left == nullptr || right == nullptr || left->val != right->val){
return false;
}
q.push(left->left);
q.push(right->right);
q.push(right->left);
q.push(left->right);
}
return true;
}
};
关键:不要去想怎么去取队列的头尾问题,转化为将头尾放在一起就变得简单容易了,对称的话需要左边取一个,右边取一个