Binary Tree Right Side View
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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 <---
For example:
Given the following binary tree,
1 <---
/ \
2 3 <---
\ \
5 4 <---
You should return [1, 3, 4]
思路:题目要求是求一个二叉树的右视图。给出的tags上面有两种方式,DFS,和BFS。
方法一:DFS
(1)右视图看过来一个树,每一层如果这个层次上有节点存在的话,只有一个节点是会被添加到最后的结果res中的。
(2)这样,res的大小实际行就代表了当前访问的层次。
(3)考虑修改先序遍历的算法,当节点不为NULL的时候,如果访问的层次大于等于res的大小,说明访问到了接下来的一层,否则则说明在这个层次上已经有右边看过来的数据添加进来,当前的元素是会被右视图遮挡的。然后优先遍历右边,而后左边,注意不同于先序遍历
代码如下:
/** * 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; helper(root,0,res); return res; } void helper(TreeNode*root,int level,vector<int>&res) { if(!root) return; if(level>=res.size()) res.push_back(root->val); helper(root->right,level+1,res); helper(root->left,level+1,res); }};方法二:BFS
使用层次遍历的方式,每一层的最后一个元素添加到结果的数组res中返回就可以
代码如下所示:
class Solution {public: vector<int> rightSideView(TreeNode* root) { vector<int> res; if(!root) return res; queue<TreeNode*> que; que.push(root); while(!que.empty()) { int size=que.size(); for(int i=0;i<size;i++) { TreeNode*front=que.front(); que.pop(); if(i==size-1) res.push_back(front->val); if(front->left) que.push(front->left); if(front->right) que.push(front->right); } } return res; }};
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- Binary Tree Right Side View
- Binary Tree Right Side View
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- Binary Tree Right Side View
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- Binary Tree Right Side View
- Binary Tree Right Side View
- Binary Tree Right Side View
- Binary Tree Right Side View
- Binary Tree Right Side View
- Binary Tree Right Side View
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