Given a binary tree, return the level order
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]
]
思路： bfs
Java代码：
import java.util.List;
import java.util.LinkedList;
/**
* 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>> resultList = new LinkedList<List<Integer>>();
if(root == null) {
return resultList;
}
List<TreeNode> arr = new LinkedList<TreeNode>();
arr.add(root);
while(!arr.isEmpty()) {
int size = arr.size();
List<Integer> list = new LinkedList<Integer>();
for(int i = 0; i < size; i++) {
TreeNode node = arr.get(0);
list.add(node.val);
if(node.left != null) arr.add(node.left);
if(node.right != null) arr.add(node.right);
arr.remove(node);
}
resultList.add(list);
}
return
