101. Symmetric Tree
Easy
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Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3]
is symmetric:
1 / \ 2 2 / \ / \ 3 4 4 3
But the following [1,2,2,null,3,null,3]
is not:
1 / \ 2 2 \ \ 3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
/**
* 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) {
if(root==NULL) return true;
vector<int> lftree;
vector<int> rgtree;
tra1(root->left,lftree);
tra2(root->right,rgtree);
//prin(lftree);
//prin(rgtree);
if(lftree!=rgtree) return false;
else return true;
}
void tra1(TreeNode* root,vector<int>& vt){//前序遍历左子树
if(root==NULL){
vt.push_back(0);
return;
}
tra1(root->left,vt);
tra1(root->right,vt);
vt.push_back(root->val);
}
void tra2(TreeNode* root,vector<int>& vt){//对称的遍历右子树
if(root==NULL){
vt.push_back(0);
return;
}
tra2(root->right,vt);
tra2(root->left,vt);
vt.push_back(root->val);
}
/*void prin(vector<int> a){
for(int i=0;i<a.size();i++){
cout<<a[i]<<' ';
}
cout<<endl;
}*/
};