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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]
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利用BFS遍历二叉树的方法,
queue<TreeNode*> q;
curr/next分别记录当前层要遍历的节点数量,下层要遍历的节点数量
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/** * 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> re; if(root==nullptr) return re; help_rightSv(root,re); for(auto i:re){ cout<<i<<" "; }cout<<endl; return re; } void help_rightSv(TreeNode *root,vector<int> &path){ queue<TreeNode*> q; int curr = 1; int next = 0; int level = 0; q.push(root); while(!q.empty()){ if(curr>0){ TreeNode *tmp = q.front(); if(path.size()==level) { path.push_back(tmp->val); } q.pop(); curr--; if(tmp->right!=nullptr){ q.push(tmp->right); next++; } if(tmp->left!=nullptr){ q.push(tmp->left); next++; } }else{ curr = next; next = 0; level++; } }///while } };
199. Binary Tree Right Side View
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原文地址:http://www.cnblogs.com/li-daphne/p/5618745.html