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 = -1;
        bfs(root, dep + 1);
        vector<int> res;
        for(int i = 0; i < ret.size(); ++i){
            res.push_back(ret[i][ret[i].size() - 1]);
        }
        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 + 1);
        if(root->right)
            bfs(root->right, depth + 1);
    }
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, 0);
        for(int i = 0; i < ret.size(); ++i){
            res.add(ret.get(i).get(ret.get(i).size() - 1));
        }
        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 + 1);
        if(root.right != null)
            bfs(ret, root.right, dep + 1);
    }
}