LeetCode101. Symmetric Tree(二叉树 / 模拟)
- Easy
- Accepted:362,120
- Submissions:847,066
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
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
Note:
Bonus points if you could solve it both recursively and iteratively.
链接
https://leetcode.com/problems/symmetric-tree
题意
给定一棵二叉树,判断它是否是对称的。
题解
之前有两道题,一道是反转一棵二叉树:LeetCode226. Invert Binary Tree(二叉树 / 模拟),还有一道是判断两颗二叉树是否相同:LeetCode100. Same Tree(二叉树 / 模拟),所以一个稍麻烦的做法是先将其反转然后判断与原二叉树是否相同。
还有一种做法就是转化为比较两颗二叉树是否相同,“左右”和“右左”相比。
同样的我们可以使用队列来进行“左右”和“右左”的比较,每次压入两个结点即可,这个思路实在是太巧妙了。不过注意一下当p和q为空时的判断。
代码
- Runtime: 12 ms, faster than 97.51% of C++ online submissions for Symmetric Tree.
- Memory Usage: 15.9 MB, less than 71.66% of C++ online submissions for Symmetric Tree.
class Solution {
public:
bool isSymmetric(TreeNode* root) {
return isSame(root, root);
}
private:
bool isSame(TreeNode* p, TreeNode* q)
{
if (p == NULL || q == NULL)
return (p == q);
if (p->val != q->val)
return false;]
return isSame(p->left, q->right) && isSame(p->right, q->left);
}
};
- Runtime: 12 ms, faster than 97.51% of C++ online submissions for Symmetric Tree.
- Memory Usage: 15.8 MB, less than 76.90% of C++ online submissions for Symmetric Tree.
class Solution {
public:
bool isSymmetric(TreeNode* root) {
queue <TreeNode*> q;
q.push(root);
q.push(root);
while (!q.empty())
{
TreeNode* p1 = q.front();
q.pop();
TreeNode* p2 = q.front();
q.pop();
if (p1 == NULL && p2 == NULL)
continue;
if (p1 == NULL || p2 == NULL)
return false;
if (p1->val != p2->val)
return false;
q.push(p1->left);
q.push(p2->right);
q.push(p1->right);
q.push(p2->left);
}
return true;
}
};
The end.
2019年3月7日 星期四