标签：style color io strong for ar art div
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.
-----SOLUTION-----
class Solution {
public:
bool isSymmetric(TreeNode *root) {
if(!root) return true;
bool result;
if((root->left==NULL && root->right != NULL) ||(root->left!=NULL && root->right == NULL))
{
return false;
}
else if(root->left == NULL && root->right == NULL)
{
return true;
}
else
{
result = cmp(root->left, root->right);
}
}
bool cmp(TreeNode * node1, TreeNode* node2)
{
int result1 = true;
int result2 = true;
if(node1->val!=node2->val) return false;
else
{
if((node1->left==NULL && node2->right != NULL) ||
(node1->left!=NULL && node2->right == NULL)||
(node1->right!=NULL && node2->left == NULL)||
(node1->right==NULL && node2->left != NULL))
{
return false;
}
if((node1->left == NULL && node2->right == NULL)&&
(node1->right == NULL && node2->left== NULL))
{
return true;
}
if(node1->left != NULL)
{
result1 = cmp(node1->left,node2->right);
}
if(node1->right != NULL)
{
result2 = cmp(node1->right,node2->left);
}
return (result1 && result2);
}
}
};
class Solution {
public:
bool isSymmetric(TreeNode *root) {
// Note: The Solution object is instantiated only once and is reused by each test case.
if(root == NULL) return true;
queue<TreeNode*> q;
q.push(root->left);
q.push(root->right);
TreeNode *t1, *t2;
while(!q.empty()){
t1 = q.front();
q.pop();
t2 = q.front();
q.pop();
if(t1 == NULL && t2 == NULL)
continue;
if(t1 == NULL || t2 == NULL || t1->val != t2->val)
return false;
q.push(t1->left);
q.push(t2->right);
q.push(t1->right);
q.push(t2->left);
}
return true;
}
};标签：style color io strong for ar art div
原文地址：http://www.cnblogs.com/mengfanrong/p/3918100.html
