/**
* 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:
bool isSymmetric(TreeNode* root) {
return isEqual(root, root);
}
private:
bool isEqual(TreeNode* left, TreeNode* right) {
if(left == NULL && right == NULL) return true;
else if(left == NULL || right == NULL) return false;
else return (left->val == right->val)
&& isEqual(left->left, right->right)
&& isEqual(left->right, right->left);
}
};
除了用 recursion 之外,我們也可以用來解,做法就是我們每次都把相對應要比較的兩個 node 放進 queue 裡面,每次 pop 出兩個 node 來比較。
/**
* 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:
bool isSymmetric(TreeNode* root) {
queue<TreeNode*> q;
q.push(root);
q.push(root);
while(!q.empty()) {
TreeNode* t1 = q.front();
q.pop();
TreeNode* t2 = q.front();
q.pop();
if(t1==NULL && t2==NULL) continue;
if(t1==NULL || t2==NULL) return false;
if(t1->val != t2->val) return false;
q.push(t1->left);
q.push(t2->right);
q.push(t1->right);
q.push(t2->left);
}
return true;
}
};