【树】199. 二叉树的右视图

题目：

给定一棵二叉树，想象自己站在它的右侧，按照从顶部到底部的顺序，返回从右侧所能看到的节点值。

示例:

输入: [1,2,3,null,5,null,4]
输出: [1, 3, 4]
解释:

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

 

解答：

方法一：BFS

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if(root == null) return res;
        
        LinkedList<TreeNode> queue = new LinkedList<>();
        queue.addLast(root);

        while(!queue.isEmpty()){
            int size = queue.size();
            int curRightView = 0;
            for(int i = 0;i<size;i++){
                TreeNode node = queue.removeFirst();
                if(i == size-1) curRightView = node.val;
                if(node.left!=null) queue.addLast(node.left);
                if(node.right!=null) queue.addLast(node.right);
            }
            res.add(curRightView);
        }
        return res;
    }
}

方法二：DFS

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    List<Integer> res = new ArrayList<>();
    public List<Integer> rightSideView(TreeNode root) {
        if(root == null) return res;
        dfs(root,0);
         
        return res;
    }

    public void dfs(TreeNode root, int level){
        if(root == null) return;

        if(level == res.size()){
            res.add(root.val);
        }
        level++;
        
        dfs(root.right,level);
        dfs(root.left,level);
    }
}