Symmetric Tree
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following is not:
1
/ \
2 2
\ \
3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
对称的递归表达式：
testMirror(leftNode->left,rightNode->right)&&testMirror(leftNode->right,rightNode->left);
leftNode->left->val==rightNode->right->val
leftNode->right->val==rightNode->left->val
class Solution {
public:
bool isSymmetric(TreeNode *root) {
if(root==NULL)
{
return true;
}
return testMirror(root->left,root->right);
}
bool testMirror(TreeNode *leftNode,TreeNode *rightNode)
{
if(leftNode==NULL&&rightNode==NULL)
return true;
if(leftNode!=NULL&&rightNode==NULL||leftNode==NULL&&rightNode!=NULL)
return false;
if(leftNode->val!=rightNode->val)
return false;
return testMirror(leftNode->left,rightNode->right)&&testMirror(leftNode->right,rightNode->left);
}
};