Java for LeetCode 102 Binary Tree Level Order Traversal

標籤：

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]]

解題思路：

前序走訪(BFS)，使用一個Queue儲存每層元素即可，JAVA實現如下：

    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;    }

 

Java for LeetCode 102 Binary Tree Level Order Traversal