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]
.
思路:BFS
class Solution { public: vector<int> rightSideView(TreeNode *root) { vector<int> ans; if(root == NULL) return ans; queue<TreeNode *> Q; Q.push(root); while(!Q.empty()) { ans.push_back(Q.front()->val); int pos = Q.size(); //当前层的元素个数 while(pos != 0) { if(Q.front()->right != NULL) Q.push(Q.front()->right); if(Q.front()->left != NULL) Q.push(Q.front()->left); Q.pop(); pos--; } } return ans; } };