LeetCode OJ：Binary Tree Right Side View（右侧视角下的二叉树）

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].

其实题目的意思就是相当于二叉树每一行的最右边的一个元素，那么简单，先bfs后取每一个数组的最后一位数就可以了，代码如下：

 /**
  * Definition for a binary tree node.
  * 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) {
         int dep = -;
         bfs(root, dep + );
         vector<int> res;
         for(int i = ; i < ret.size(); ++i){
             res.push_back(ret[i][ret[i].size() - ]);
         }
         return res;
     }

     void bfs(TreeNode * root, int depth)
     {
         if(root == NULL) return;
         if(depth < ret.size()){
             ret[depth].push_back(root->val);
         }else{
             vector<int>tmp;
             ret.push_back(tmp);
             ret[depth].push_back(root->val);
         }
         if(root->left)
             bfs(root->left, depth + );
         if(root->right)
             bfs(root->right, depth + );
     }
 private:
     vector<vector<int>> ret;
 };

java版本的代码如下所示，同样是BFS之后再取最后一位组成一个数组：

 public class Solution {
     public List<Integer> rightSideView(TreeNode root) {
         List<List<Integer>> ret = new ArrayList<List<Integer>>();
         List<Integer> res = new ArrayList<Integer>();
         if(root == null)
             return res;
         bfs(ret, root, );
         for(int i = ; i < ret.size(); ++i){
             res.add(ret.get(i).get(ret.get(i).size() - ));
         }
         return res;
     }

     public void bfs(List<List<Integer>> ret, TreeNode root, int dep){
         if(ret.size() <= dep){
             ret.add(new ArrayList<Integer>());
         }
         ret.get(dep).add(root.val);
         if(root.left != null)
             bfs(ret, root.left, dep + );
         if(root.right != null)
             bfs(ret, root.right, dep + );
     }
 }