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]
.
Credits:
分析:层次遍历,取最右面的元素:
方法一:双q法:
class Solution { public: vector<int> rightSideView(TreeNode *root) { queue<TreeNode*> q1; queue<TreeNode*> q2; vector<int> res; if(root != NULL) { q1.push(root); } while(!q1.empty()) { TreeNode * p = q1.front(); q1.pop(); if(p->left) q2.push(p->left); if(p->right) q2.push(p->right); if(q1.empty() /*&& !q2.empty()*/) { res.push_back(p->val); swap(q1, q2); } } return res; } };
方法二:单q+ NULL标记level结束法
/** * 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> rtn; if(root == NULL) return rtn; queue<TreeNode*> q; TreeNode* pre = NULL; q.push(root); q.push(NULL);//mark the end of this level while(!q.empty()) { TreeNode * p = q.front(); q.pop(); if(p == NULL) { rtn.push_back(pre->val); if(!q.empty())// some elements still exist in q. q.push(NULL);//mark end of this level } else { if(p->left) q.push(p->left); if(p->right) q.push(p->right); } pre = p; } return rtn; } };