[GeeksForGeeks] Print leftmost and rightmost nodes at each level of a binary tree.

最新推荐文章于 2025-12-22 15:57:21 发布

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原文链接：http://www.cnblogs.com/lz87/p/7277712.html

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#java #开发工具

本文介绍了一种使用队列实现的二叉树层级遍历算法，并特别关注每一层的最左侧和最右侧节点。通过Java实现的具体示例，展示了如何获取这些节点。

Given a Binary Tree, Print the corner nodes at each level. The node at the leftmost and the node at the rightmost.

For example, output for following is 15, 10, 20, 8, 25.

Solution. Level Order Traversal using queue.

Core idea: Level order traversal always visit nodes of one level from left to right.

And we know the number of nodes at each level by pre-reading the size of the queue.

 1 import java.util.ArrayList;
 2 import java.util.LinkedList;
 3 import java.util.Queue;
 4 
 5 class TreeNode {
 6     TreeNode left;
 7     TreeNode right;
 8     int val;
 9     TreeNode(int val){
10         this.left = null;
11         this.right = null;
12         this.val = val;
13     }
14 }
15 public class Solution {
16     public ArrayList<TreeNode> getLeftRightMostAtEachLevel(TreeNode root) {
17         ArrayList<TreeNode> result = new ArrayList<TreeNode>();
18         if(root == null){
19             return result;
20         }
21         Queue<TreeNode> queue = new LinkedList<TreeNode>();
22         queue.offer(root);        
23         while(!queue.isEmpty()){
24             int size = queue.size();
25             for(int i = 0; i < size; i++){
26                 TreeNode curr = queue.poll();
27                 if(i == 0){
28                     result.add(curr);
29                 }
30                 if(i > 0 && i == size - 1){
31                     result.add(curr);
32                 }
33                 if(curr.left != null){
34                     queue.offer(curr.left);
35                 }
36                 if(curr.right != null){
37                     queue.offer(curr.right);
38                 }
39             }
40         }
41         return result;
42     }
43 }