199. Binary Tree Right Side View
Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.
Example:
Input: [1,2,3,null,5,null,4] Output: [1, 3, 4] Explanation: 1 <--- / \ 2 3 <--- \ \ 5 4 <---
给定一棵二叉树,问从树的右边往左看,从上往下,看到的数组是什么。很有意思哈,一棵树变着花样出题。
想到的话,其实很简单,从右边看到的数是每一层最右边的数(好像是废话),所以我们可以对树层次遍历,取出每层最右边的数,构成的数组就是答案。完整代码如下:
class Solution {
public:
vector<int> rightSideView(TreeNode* root)
{
vector<int> ans;
if (root == NULL)
return ans;
queue<TreeNode*> q;
q.push(root);
while (!q.empty()) {
ans.push_back(q.back()->val);
int n = q.size();
for (int i = 0; i < n; ++i) {
TreeNode* tmp = q.front();
q.pop();
if (tmp->left)
q.push(tmp->left);
if (tmp->right)
q.push(tmp->right);
}
}
return ans;
}
};
本代码提交AC,用时3MS。