[leetcode]Binary Tree Level Order Traversal

最新推荐文章于 2025-08-19 14:43:05 发布

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最新推荐文章于 2025-08-19 14:43:05 发布

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分类专栏：算法文章标签：算法

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本文介绍了一种算法，用于实现二叉树的层次遍历并返回各层节点的值。通过使用栈和列表结构，该算法能有效地按从左到右、由上至下的顺序收集节点值。

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问题描述：

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? > read more on how binary tree is serialized on OJ.

代码：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {  //java
		if(root == null)
			return new ArrayList<>();
			
		Stack<TreeNode> stack = new Stack<TreeNode>();
		stack.add(root);
		List<TreeNode> levelList = new ArrayList<TreeNode>();
		
		
		List<List<Integer>> result = new ArrayList<List<Integer>>();
		List<List<Integer>> invert_result = new ArrayList<List<Integer>>();
		List<Integer> tmp = new ArrayList<Integer>();
		
		while(!stack.isEmpty()|| !levelList.isEmpty()){
			if(stack.isEmpty()){
				for(int i=levelList.size()-1; i>=0; i--)
					stack.push(levelList.get(i));
				levelList.clear();
				invert_result.add(tmp);
				tmp = new ArrayList<Integer>();
			}
			
			while(!stack.isEmpty()){
				TreeNode node = stack.pop();
				tmp.add(node.val);
				
				if(node.left !=null)
					levelList.add(node.left);
				if(node.right !=null)
					levelList.add(node.right);
			}
		}
		invert_result.add(tmp);
		
	
		
		return invert_result;
        
    }
}