Symmetric Tree対称ツリー
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
But the following is not:
1.私の回答(再帰)
2.非再帰
For example, this binary tree is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following is not:
1
/ \
2 2
\ \
3 31.私の回答(再帰)
/**
* 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 isSysm(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 && isSysm(left->right,right->left) && isSysm(left->left, right->right);
}
}
bool isSymmetric(TreeNode* root) {
if(!root)
return true;
if(root->left==NULL && root->right==NULL)
return true;
TreeNode* left = root->left;
TreeNode* right = root->right;
return isSysm(left,right);
}
};2.非再帰
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
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;
}
};