Java for LeetCode 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,#,#,15,7},

return its level order traversal as:

[

  [3],

  [9,20],

  [15,7]

]

解题思路：

前序遍历(BFS)，使用一个Queue存储每层元素即可，JAVA实现如下：

    public List<List<Integer>> levelOrder(TreeNode root) {

        List<List<Integer>> list = new ArrayList<List<Integer>>();

		if (root == null)

			return list;

		Queue<TreeNode> queue = new LinkedList<TreeNode>();

		queue.add(root);

		while (queue.size() != 0) {

			List<Integer> alist = new ArrayList<Integer>();

			for (TreeNode child : queue)

				alist.add(child.val);

			list.add(new ArrayList<Integer>(alist));

			Queue<TreeNode> queue2=queue;

			queue=new LinkedList<TreeNode>();

			for(TreeNode child:queue2){

				if (child.left != null)

					queue.add(child.left);

				if (child.right != null)

					queue.add(child.right);

			}

		}

		return list;

    }

巴特西

Java for LeetCode 102 Binary Tree Level Order Traversal

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