[LeetCode] Binary Tree Right Side View
2015-04-06 17:00
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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].
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); } };
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