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.
For example: Given the following binary tree,
1 <--- / \ 2 3 <--- \ \ 5 4 <---
You should return [1, 3, 4].
解法一：
类似level order tree traversal，用一个queue存储每一层的node，将每一层的最后一个node，也就是最右的node推到结果中。
/** * 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> res; if(!root) return res; queue<TreeNode*> q; q.push(root); while(!q.empty()){ int len = q.size(); for(int i=0; i<len; i++){ TreeNode* node = q.front(); q.pop(); if(node->left) q.push(node->left); if(node->right) q.push(node->right); if(i==len-1) res.push_back(node->val); } } return res; } };
