[LeetCode] Level Order Traversal

题目说明

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]

]

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}".

思路

这道题可以这么做：用一个队列来保存每层节点，遍历该层所有节点同时将每个节点加到结果中，同时将每个节点的子女节点用一个list保存下来，遍历完本层节点后，将保存的子女节点加到队列中继续遍历。直到子女节点为空（也就是队列为空）为止

代码

public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
            // Note: The Solution object is instantiated only once and is reused by each test case.
            ArrayList<ArrayList<Integer>> ans=new ArrayList<ArrayList<Integer>>();
            if(root==null)
                return ans;
            Queue<TreeNode> list=new LinkedList<TreeNode>();
            list.add(root);
            while(!list.isEmpty())
            {
                ArrayList<TreeNode> levelNodes=new ArrayList<TreeNode>();
                ArrayList<Integer> res=new ArrayList<Integer>();
                while(!list.isEmpty())
                {
                    TreeNode node=list.poll();
                    if(node.left!=null)
                        levelNodes.add(node.left);
                    if(node.right!=null)
                        levelNodes.add(node.right);
                    res.add(node.val);
                }
                list.addAll(levelNodes);
                ans.add(res);
            }
            return ans;
        }
    }