1 // 我的代码
2 package Leetcode;
3 /**
4 * 199. Binary Tree Right Side View
5 * address: https://leetcode.com/problems/binary-tree-right-side-view/
6 * Given a binary tree, imagine yourself standing on the right side of it,
7 * return the values of the nodes you can see ordered from top to bottom.
8 *
9 */
10 import java.util.*;
11
12 public class BinaryTreeRightSideView {
13 public static void main(String[] args) {
14 TreeNode[] tree = new TreeNode[9];
15 for (int i = 1; i < 9; i++){
16 tree[i] = new TreeNode(i);
17 }
18 TreeNode.link(tree, 1, 2, 3);
19 TreeNode.link(tree, 2, 4, 5);
20 TreeNode.link(tree, 3, 6, -1);
21 TreeNode.link(tree, 5, 7, 8);
22 // 先序遍历
23 TreeNode.prePrint(tree[1]);
24 // solution
25 List<Integer> result = rightSideView(tree[1]);
26 System.out.println(result);
27
28 // solution2
29 List<Integer> result2 = rightSideView2(tree[1]);
30 System.out.println(result2);
31
32 }
33 /**
34 * 方法一：时间复杂度较高，需要 O(n),n为树中节点的个数
35 * @param root
36 * @return
37 */
38 public static List<Integer> rightSideView(TreeNode root) {
39 List<Integer> result = new ArrayList<>();
40 Queue<TreeNode> queue = new LinkedList<>();
41 int childNum = 0;
42 if (root != null){
43 queue.offer(root);
44 while(!queue.isEmpty()){
45 TreeNode node = queue.poll();
46 if(node.left!= null){
47 queue.offer(node.left);
48 childNum++;
49 }
50 if (node.right != null)
51 {
52 queue.offer(node.right);
53 childNum++;
54 }
55 System.out.println("queue.size() = " + queue.size());
56 if (childNum - queue.size() == 0){
57 result.add(node.val);
58 childNum = 0;
59 }
60 }
61 }
62
63 return result;
64
65 }
1 /**
2 * 方法二
3 * 别人家的解法，巧妙，速度快且好理解
4 * @param root
5 * 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
6 * @return
7 */
8 public static List<Integer> rightSideView2(TreeNode root) {
9 List<Integer> res = new ArrayList<Integer>();
10 if (root == null){
11 return res;
12 }
13 dfs (root, res, 0);
14 return res;
15 }
16
17 public static void dfs (TreeNode root, List<Integer> res, int level){
18 if (root == null){
19 return;
20 }
21 if (res.size() == level){
22 res.add (root.val);
23 }
24 if (root.right != null){
25 dfs (root.right, res, level + 1);
26 }
27 if (root.left != null){
28 dfs (root.left, res, level + 1);
29 }
30 }
