【Leetcode】Binary Tree Level Order Traversal I

最新推荐文章于 2025-09-07 20:32:33 发布

yyz20002008

最新推荐文章于 2025-09-07 20:32:33 发布

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文章标签： java leetcode arraylist LinkedList TreeNode

本文链接：https://blog.youkuaiyun.com/yyz20002008/article/details/38344641

本文介绍了一种解决二叉树层次遍历问题的方法，并提供了两种不同的Java实现方案。一种使用了两个链表来记录每一层的节点，另一种则利用队列实现了层次遍历。

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

confused what "{1,#,2,3}" means?

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

Java:

1.同学讲解的解法：

用两个链表操作

public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {  
       ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
       if(root==null) return res;
       ArrayList<TreeNode> level=new ArrayList<TreeNode>();
       level.add(root);
       while(level.size()!=0)
       {
           ArrayList<TreeNode> next=new ArrayList<TreeNode>();
           ArrayList<Integer> list= new ArrayList<Integer>();
           for(TreeNode node:level){
               list.add(node.val);
               if(node.left!=null)
               {
                   next.add(node.left);
               }
               if(node.right!=null)
               {
                   next.add(node.right);
               }
           }
           res.add(list);
           level=next;
       }
       return res;  
    }  
}

其他解法：

http://blog.youkuaiyun.com/linhuanmars/article/details/23404111

征服代码的也好懂

public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {  
    ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();  
    if(root == null)  
        return res;  
    LinkedList<TreeNode> queue = new LinkedList<TreeNode>();  
    queue.add(root);  
    int curNum = 0;  
    int lastNum = 1;  
    ArrayList<Integer> list = new ArrayList<Integer>();  
    while(!queue.isEmpty())  
    {  
        TreeNode cur = queue.poll();  
        lastNum--;  
        list.add(cur.val);  
        if(cur.left!=null)  
        {  
            queue.add(cur.left);  
            curNum ++;  
        }  
        if(cur.right!=null)  
        {  
            queue.add(cur.right);  
            curNum++;  
        }  
        if(lastNum==0)  
        {  
            lastNum = curNum;  
            curNum = 0;  
            res.add(list);  
            list = new ArrayList<Integer>();  
        }  
    }  
    return res;  
}