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Question：
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.
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1 vector<int> rightSideView(TreeNode* root) 2 { 3 queue<TreeNode*> qtree; 4 vector<int> result; 5 vector<int> line_nodes; 6 TreeNode *visit = NULL; 7 int line_size = 1; 8 int next_line_size = 0; 9 10 if (root == NULL) 11 return result; 12 13 qtree.push(root); 14 while (!qtree.empty()) 15 { 16 visit = qtree.front(); 17 qtree.pop(); 18 line_size--; 19 line_nodes.push_back(visit->val); 20 21 if (visit->left) 22 { 23 qtree.push(visit->left); 24 next_line_size++; 25 } 26 if (visit->right) 27 { 28 qtree.push(visit->right); 29 next_line_size++; 30 } 31 32 if (line_size == 0) 33 { 34 line_size = next_line_size; 35 next_line_size = 0; 36 result.push_back(line_nodes.back()); 37 line_nodes.clear(); 38 } 39 } 40 41 return result; 42 }
由于不需要返回每层的节点，可以如下优化：
1 vector<int> rightSideView(TreeNode* root) 2 { 3 queue<TreeNode*> qtree; 4 vector<int> result; 5 TreeNode *visit = NULL; 6 7 if (root == NULL) 8 return result; 9 10 qtree.push(root); 11 while (!qtree.empty()) 12 { 13 visit = qtree.back(); 14 result.push_back(visit->val); 15 int size = qtree.size(); 16 for (int i = 0; i < size; i++) 17 { 18 visit = qtree.front(); 19 qtree.pop(); 20 if (visit->left) 21 qtree.push(visit->left); 22 if (visit->right) 23 qtree.push(visit->right); 24 } 25 } 26 27 return result; 28 }
[leetcode 199]Binary Tree Right Side View
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原文地址：http://www.cnblogs.com/ym65536/p/5004567.html
