解答：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 
 
*/
//BFS：層序遍歷保留每一層的最後一個節點
class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if(root == null) {
            return res;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        //入隊用offer
 

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

//DFS：深度搜索，每一次的第一個加入
class Solution {
    public List<Integer> res = new ArrayList<>();
    public List<Integer> rightSideView(TreeNode root) {
        dfs(root, 0);
        return res;
    }
    public void dfs(TreeNode root, int depth) {
        if(root == null) {
            return;
        }
        if(depth == res.size()) {
            res.add(root.val);
        }
        depth++;
        dfs(root.right, depth);
        dfs(root.left, depth);
    }
}

本文來自部落格園，作者：劉金輝，轉載請註明原文連結：https://www.cnblogs.com/liujinhui/p/15646842.html