题目来源：https://leetcode.com/problems/binary-tree-level-order-traversal/
问题描述
Medium
Given a binary tree, return the level ordertraversal 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]
]
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题意
二叉树宽度优先遍历
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思路
队列。
由于不同深度的值要在不同的数组中输出，所以用了队列中用的元素用了level来维护节点的深度。
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代码
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { class ListNode { int val, level; TreeNode left, right; public ListNode (int val, int level, TreeNode left, TreeNode right) { this.val = val; this.level = level; this.left = left; this.right = right; } } public List<List<Integer>> levelOrder(TreeNode root) { if (root == null) { return Collections.emptyList(); } LinkedList<ListNode> queue = new LinkedList<ListNode>(); queue.add(new ListNode(root.val, 0, root.left, root.right)); int curLevel = 0; List<List<Integer>> ret = new LinkedList<>(); List<Integer> levelList = new LinkedList<Integer>(); while (!queue.isEmpty()) { ListNode head = queue.pollFirst(); if (head.left != null) { queue.add(new ListNode(head.left.val, head.level+1, head.left.left, head.left.right)); } if (head.right != null) { queue.add(new ListNode(head.right.val, head.level+1, head.right.left, head.right.right)); } if (head.level > curLevel) { ret.add(levelList); levelList = new LinkedList<Integer>(); curLevel = head.level; } levelList.add(head.val); } ret.add(levelList); return ret; } }
