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:
Special thanks to @amrsaqr for adding this problem and creating all test cases.
Analyse: the same as Binary Tree Level Order Traversal.
1. Recursion.
Runtime: 4ms.
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<int> rightSideView(TreeNode *root) { 13 vector<int> result; 14 levelOrder(root, result, 0); 15 return result; 16 } 17 void levelOrder(TreeNode* root, vector<int>& result, int level){ 18 if(!root) return; 19 if(level == result.size()) //the level does not exist and need to create it 20 result.push_back(0); 21 result[level] = root->val; //the recursion process ensures that the right most node value is added 22 23 levelOrder(root->left, result, level + 1); 24 levelOrder(root->right, result, level + 1); 25 } 26 };
2. Iteration: When it reaches the end of the current level, push its value into the result vector.
Runtime: 4ms.
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<int> rightSideView(TreeNode *root) { 13 vector<int> result; 14 if(!root) return result; 15 16 queue<TreeNode* > qu; 17 qu.push(root); 18 //result.push_back(root->val); 19 while(!qu.empty()){ 20 int n = qu.size(); 21 while(n--){ 22 TreeNode* temp = qu.front(); 23 qu.pop(); 24 if(temp->left) qu.push(temp->left); 25 if(temp->right) qu.push(temp->right); 26 27 if(n == 0) result.push_back(temp->val); 28 } 29 } 30 return result; 31 } 32 };