102. Binary Tree Level Order Traversal

最新推荐文章于 2021-12-07 15:29:05 发布

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最新推荐文章于 2021-12-07 15:29:05 发布

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文章标签： java 数据结构与算法 runtime

原文链接：https://segmentfault.com/a/1190000018176652

本文介绍了一种使用队列实现的二叉树层次遍历算法，通过实例展示了如何从左到右，按层次获取二叉树节点的值。算法在在线提交中表现优秀，运行时间为1ms，比86.63%的Java提交更快，内存使用为37.5MB，优于所有Java提交。

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,null,null,15,7],
    3
   / \
  9  20
    /  \
   15   7
return its level order traversal as:
[
  [3],
  [9,20],
  [15,7]
]

难度：medium

题目：给定二叉树，返回其层次遍历结点值。

思路：队列(FIFO)

Runtime: 1 ms, faster than 86.63% of Java online submissions for Binary Tree Level Order Traversal.
Memory Usage: 37.5 MB, less than 100.00% of Java online submissions for Binary Tree Level Order Traversal.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (null == root) {
            return result;
        }
        
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        int currentLevelCount = 1, nextLevelCount = 0;
        List<Integer> currentLevelElements = new ArrayList<>();
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            currentLevelElements.add(node.val);
            if (node.left != null) {
                queue.add(node.left);
                nextLevelCount++;
            }
            if (node.right != null) {
                queue.add(node.right);
                nextLevelCount++;
            }
            currentLevelCount--;
            if (0 == currentLevelCount) {
                result.add(currentLevelElements);
                currentLevelElements = new ArrayList<>();
                currentLevelCount = nextLevelCount;
                nextLevelCount = 0;
            }
        }
        
        return result;
    }
}