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);
         }
     }