102. 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).

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]
]

 题目含义:按照从左到右的顺序,输出每一层的节点

 1     public List<List<Integer>> levelOrder(TreeNode root) {
 2         List<List<Integer>> result = new ArrayList<>();
 3         if (root == null) return result;
 4         Queue<TreeNode> q = new LinkedList<>();
 5         q.add(root);
 6         while (!q.isEmpty())
 7         {
 8             int size = q.size();
 9             List<Integer> levels = new ArrayList<>();
10             for (int i=0;i<size;i++)
11             {
12                 TreeNode node = q.poll();
13                 levels.add(node.val);
14                 if (node.left !=null) q.offer(node.left);
15                 if (node.right !=null) q.offer(node.right);
16             }
17             result.add(levels);
18         }
19         return result;        
20     }
 
原文地址:https://www.cnblogs.com/wzj4858/p/7715443.html