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
class Solution {
public:
bool isSymmetric(TreeNode* root) {
if (root == NULL)
return true;
vector<TreeNode*>que;
que.push_back(root);
while (!que.empty())
{
vector<TreeNode*>newque;
for (int i = 0; i < que.size(); i++)
{
if (i < que.size() / 2)
{
if (que[i] == NULL&&que[que.size() - 1 - i] != NULL)
return false;
if (que[i] != NULL&&que[que.size() - 1 - i] == NULL)
return false;
if (que[i] != NULL&&que[que.size() - 1 - i] != NULL&&que[i]->val != que[que.size() - 1 - i]->val)
return false;
}
if (que[i] != NULL)
{
newque.push_back(que[i]->right);
newque.push_back(que[i]->left);
}
}
que = newque;
}
return true;
}
};accept

本文介绍了一种算法来判断给定的二叉树是否是对称的,即树的左右子树是否镜像对称。
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