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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},
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
二叉树的广度优先搜索,用队列来实现,借助两个变量来记录每一层的节点个数,以及何时将每层的节点的值加入到结果集中。代码如下:
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
二叉树的广度优先搜索,用队列来实现,借助两个变量来记录每一层的节点个数,以及何时将每层的节点的值加入到结果集中。代码如下:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> result = new ArrayList<List<Integer>>(); List<Integer> list = new ArrayList<Integer>(); Queue<TreeNode> queue = new LinkedList<TreeNode>(); if(root == null) return result; int count = 0; int helper = 1; queue.offer(root); while(!queue.isEmpty()) { TreeNode node = queue.poll(); list.add(node.val); helper --; if(node.left != null) { queue.offer(node.left); count ++; } if(node.right != null) { queue.offer(node.right); count ++; } if(helper == 0) { result.add(new ArrayList<Integer>(list)); helper = count; count = 0; list.clear(); } } return result; } }
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