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[LeetCode] Binary Tree Right Side View

时间:2015-04-06 17:17:22      阅读:109      评论:0      收藏:0      [点我收藏+]

标签:leetcode   c++   

Binary Tree Right Side View

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].

解题思路:

这道题较为简单,可以用类似于先序遍历的办法来做,但是有点不同,遍历的顺序是根,右,左。用一个map记录每一层第一个出现的值。下面是代码:
/**
 * Definition for binary tree
 * 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) {
        map<int, int> levelToVal;   //记录每一层第一个出现的数值。
        tranver(root, 0, levelToVal);
        
        vector<int> result;
        
        for(map<int, int>::iterator it=levelToVal.begin(); it!=levelToVal.end(); it++){
            result.push_back(it->second);
        }
        
        return result;
    }
    
    //类似于先序遍历,但是,顺序是上,右,左
    void tranver(TreeNode* node, int level, map<int, int>& levelToVal){
        if(node==NULL){
            return;
        }
        map<int, int>::iterator it=levelToVal.find(level);
        if(it==levelToVal.end()){
            levelToVal.insert(map<int, int>::value_type(level, node->val));
        }
        tranver(node->right, level+1, levelToVal);
        tranver(node->left, level+1, levelToVal);
    }
};


[LeetCode] Binary Tree Right Side View

标签:leetcode   c++   

原文地址:http://blog.csdn.net/kangrydotnet/article/details/44903369

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