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LevelOrder.java
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package ds;
import java.util.*;
/**
* 给定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。
* <p>
* 例如:
* 给定二叉树: [3,9,20,null,null,15,7],
* <p>
* 3
* / \
* 9 20
* / \
* 15 7
* 返回其层次遍历结果：
* <p>
* [
* [3],
* [9,20],
* [15,7]
* ]
*
* @author yangyi 2019年01月28日11:10:18
*/
public class LevelOrder {
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
}
public List<List<Integer>> levelOrder(TreeNode root) {
if (root == null) {
return new LinkedList<>();
}
List<List<Integer>> result = new LinkedList<>();
Queue<TreeNode> queue = new ArrayDeque<>();
queue.offer(root);
while (!queue.isEmpty()) {
//标尺队列的长度，这里一定要用一个指针指针指向一下，否则就会出错
int levelSize = queue.size();
List<Integer> integerList = new ArrayList<>();
for (int i = 0; i < levelSize; i++) {
TreeNode node = queue.poll();
if (node != null) {
integerList.add(node.val);
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
}
result.add(integerList);
}
return result;
}
public TreeNode createTree() {
TreeNode node_3 = new TreeNode();
node_3.val = 3;
TreeNode node_9 = new TreeNode();
node_9.val = 9;
TreeNode node_20 = new TreeNode();
node_20.val = 20;
TreeNode node_15 = new TreeNode();
node_15.val = 15;
TreeNode node_7 = new TreeNode();
node_7.val = 7;
node_3.left = node_9;
node_3.right = node_20;
node_9.left = node_15;
node_20.right = node_7;
return node_3;
}
public static void main(String[] args) {
LevelOrder levelOrder = new LevelOrder();
List<List<Integer>> result = levelOrder.levelOrder(levelOrder.createTree());
System.out.println("遍历结果出来看看:");
for (List<Integer> integerList : result) {
for (Integer integer : integerList) {
System.out.println(integer);
}
}
}
}