题目描述:
给定一棵二叉树,想象自己站在它的右侧,按照从顶部到底部的顺序,返回从右侧所能看到的节点值。
示例:
输入: [1,2,3,null,5,null,4] 输出: [1, 3, 4] 解释: 1 <--- / \ 2 3 <--- \ \ 5 4 <---
思路:使用一个队列,做层次遍历,每次都把下一层的节点保存在这个队列里面,那么这个队列最右边的节点就是站在右边可以看到的节点
代码:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> rightSideView(TreeNode* root) {
vector<int> ret;
if(root == NULL)
return ret;
queue<TreeNode* > next_q;
if(root->left != NULL)
next_q.push(root->left);
if(root->right != NULL)
next_q.push(root->right);
ret.push_back(root->val);
while(!next_q.empty()){
ret.push_back(next_q.back()->val);
queue<TreeNode* > temp_q;
while(!next_q.empty()){
TreeNode* tempNode = next_q.front();
if(tempNode->left != NULL)
temp_q.push(tempNode->left);
if(tempNode->right != NULL)
temp_q.push(tempNode->right);
next_q.pop();
}
next_q = temp_q;
}
return ret;
}
};