101. Symmetric Tree
2016-08-24 19:08
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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
But the following
/**
* 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 testSymmetric(TreeNode* left, TreeNode* right) {
if(left==NULL&&right==NULL)
return true;
else if(left==NULL&&right!=NULL)
return false;
else if(left!=NULL&&right==NULL)
return false;
else if(left!=NULL&&right!=NULL&&left->val!=right->val)
return false;
else
return testSymmetric(left->left,right->right)&&testSymmetric(left->right,right->left);
}
bool isSymmetric(TreeNode* root) {
if (root==NULL)
return true;
else
return testSymmetric(root->left,root->right);
}
};
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
/**
* 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 testSymmetric(TreeNode* left, TreeNode* right) {
if(left==NULL&&right==NULL)
return true;
else if(left==NULL&&right!=NULL)
return false;
else if(left!=NULL&&right==NULL)
return false;
else if(left!=NULL&&right!=NULL&&left->val!=right->val)
return false;
else
return testSymmetric(left->left,right->right)&&testSymmetric(left->right,right->left);
}
bool isSymmetric(TreeNode* root) {
if (root==NULL)
return true;
else
return testSymmetric(root->left,root->right);
}
};
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