leetcode 199 :Binary Tree Right Side View

// 我的代码
package Leetcode;
/**
 * 199. Binary Tree Right Side View
 * address: https://leetcode.com/problems/binary-tree-right-side-view/
 * 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.
 * 
 */
import java.util.*;

public class BinaryTreeRightSideView {
    public static void main(String[] args) {
        TreeNode[] tree = new TreeNode[9];
        for (int i = 1; i < 9; i++){
            tree[i] = new TreeNode(i);
        }
        TreeNode.link(tree, 1, 2, 3);
        TreeNode.link(tree, 2, 4, 5);
        TreeNode.link(tree, 3, 6, -1);
        TreeNode.link(tree, 5, 7, 8);
        // 先序遍历
        TreeNode.prePrint(tree[1]);
        // solution
        List<Integer> result = rightSideView(tree[1]);
        System.out.println(result);
        
        // solution2
        List<Integer> result2 = rightSideView2(tree[1]);
        System.out.println(result2);
        
    }
    /**
     * 方法一：时间复杂度较高，需要 O(n),n为树中节点的个数
     * @param root
     * @return
     */
    public static List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        Queue<TreeNode> queue = new LinkedList<>();
        int childNum = 0;
            if (root != null){
                queue.offer(root);
                while(!queue.isEmpty()){
                    TreeNode node = queue.poll(); 
                    if(node.left!= null){
                        queue.offer(node.left);
                        childNum++;
                    }
                    if (node.right != null)
                    {
                        queue.offer(node.right);
                        childNum++;
                    }
                    System.out.println("queue.size() = " + queue.size());
                    if (childNum - queue.size() == 0){
                        result.add(node.val);
                        childNum = 0;
                    }
                } 
        }

        return result;
        
    }

/**
     * 方法二
     * 别人家的解法，巧妙，速度快且好理解
     * @param root 
     * 中右左 深度优先遍历，保存每层的第一个节点。用当前结果表的size作为标识,这样保证每次存的节点都是第一次出现在该层的节点
     * @return
     */
    public   static List<Integer> rightSideView2(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if (root == null){
            return res;
        }
        dfs (root, res, 0);
        return res;
    }
    
    public static void dfs (TreeNode root, List<Integer> res, int level){
        if (root == null){
            return;
        }
        if (res.size() == level){
            res.add (root.val);
        }
        if (root.right != null){
            dfs (root.right, res, level + 1);
        }
        if (root.left != null){
            dfs (root.left, res, level + 1);
        }
    }