2020-01-16 09:50:29 +08:00

98 lines
2.4 KiB
Java

// Source : https://oj.leetcode.com/problems/binary-tree-level-order-traversal/
// Inspired by : http://www.jiuzhang.com/solutions/binary-tree-level-order-traversal/
// Author : Lei Cao
// Date : 2015-10-08
/**********************************************************************************
*
* 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]
* ]
*
* confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
*
* OJ's Binary Tree Serialization:
*
* The serialization of a binary tree follows a level order traversal, where '#' signifies
* a path terminator where no node exists below.
*
* Here's an example:
*
* 1
* / \
* 2 3
* /
* 4
* \
* 5
*
* The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".
*
*
**********************************************************************************/
package binaryTreeLevelOrderTraversal;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
public class binaryTreeLevelOrderTraversal {
/**
* @param root: The root of binary tree.
* @return: Level order a list of lists of integer
*/
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> results = new ArrayList<>();
if (root == null) {
return results;
}
ArrayList<Integer> values = new ArrayList<Integer>();
Queue<TreeNode> q = new LinkedList<>();
q.offer(root);
q.offer(null);
while (q.size() > 0) {
TreeNode node = q.poll();
// null node used as a separator of every level
if (node == null) {
results.add(new ArrayList<>(values));
values.clear();
if (q.size() == 0) {
break;
}
q.offer(null);
continue;
}
values.add(node.val);
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
return results;
}
}