LintCode 69. Binary Tree Level Order Traversal 原创Java参考解答
问题描述
http://www.lintcode.com/en/problem/binary-tree-level-order-traversal/
Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).
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] ]
解题思路
题目求做二叉树的层序遍历。
二叉树层序遍历是广度优先搜索BFS的最常见应用。
- 异常检查root节点是否为空
- 设一个装每一层node的queue队列,首先把root节点放入队列中。
- 把队列中node一层一层放入动态数组level中。并通过node.left 和 node.right把下一层的nodes放入queue中,以便下一次循环继续进行。把level结果加入result中。
算法的复杂度是就节点的数量,O(N),空间复杂度是一层的节点数,也是O(N)。
参考代码
/** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /** * @param root: The root of binary tree. * @return: Level order a list of lists of integer */ public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) { ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); if (root == null) { return result; } Queue<TreeNode> queue = new LinkedList<>(); queue.offer(root); while (!queue.isEmpty()) { int size = queue.size(); ArrayList<Integer> level = new ArrayList<Integer>(); for (int i = 0; i < size; i++) { TreeNode node = queue.poll(); level.add(node.val); if (node.left != null) { queue.offer(node.left); } if (node.right != null) { queue.offer(node.right); } } result.add(level); } return result; } }