199. Binary Tree Right Side View ----M

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.

Example:

Input: [1,2,3,null,5,null,4]
Output: [1, 3, 4]
Explanation:

   1            <---
 /   \
2     3         <---
 \     \
  5     4       <---

A:
 我所能想到的，就是简单的bfs (dfs)

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> list = new LinkedList<Integer>();
        Queue<TreeNode> queue = new LinkedList();
        if(root == null)
            return list;
        queue.add(root);
        while( queue.isEmpty() == false){
            Queue<TreeNode> newQueue = new LinkedList();
            boolean isFirst = true;
            
            while( ! queue.isEmpty()){
                TreeNode node = queue.poll();
                if(isFirst){
                    list.add(node.val);
                    isFirst = false;
                }
                if(node.right != null)
                    newQueue.add(node.right);
                if(node.left != null)  
                    newQueue.add(node.left);
            }
            queue = newQueue;
        }
        return list;
    }
}

***********dfs ****************

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int> res;
        helper(root, 1, res);
        return res;
    }
private:
    void helper(TreeNode * root, int depth, vector<int> & res){
        if(!root)
            return;
        if(res.size() < depth){
            res.push_back(root->val);
        }
        helper(root->right, depth+1, res);
        helper(root->left, depth+1, res);
    }
};

Mistakes:



Learned: