Leetcode 199: Binary Tree Right Side View

最新推荐文章于 2020-04-04 21:56:32 发布

转载最新推荐文章于 2020-04-04 21:56:32 发布 · 56 阅读

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原文链接：http://www.cnblogs.com/liangmou/p/7917981.html

本文介绍了一种算法，该算法可以遍历一棵二叉树并返回从顶部到底部的右侧可见节点值。通过使用队列进行层级遍历，确保了只收集每层最右侧的节点值。

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.

For example:
Given the following binary tree,

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

You should return [1, 3, 4].

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public int val;
 5  *     public TreeNode left;
 6  *     public TreeNode right;
 7  *     public TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public IList<int> RightSideView(TreeNode root) {
12         var queue = new Queue<TreeNode>();
13         var result = new List<int>();
14         
15         if (root == null) return result;
16         
17         queue.Enqueue(root);
18         
19         int count = 1;
20         while (queue.Count > 0)
21         {
22             for (int i = 0; i < count; i++)
23             {
24                 var n = queue.Dequeue();
25                 if (n.left != null) queue.Enqueue(n.left);
26                 if (n.right != null) queue.Enqueue(n.right);
27                 
28                 if (i == count - 1) result.Add(n.val);
29             }
30             
31             count = queue.Count;
32         }
33         
34         return result;
35     }
36 }