LeetCode-Binary Tree Level Order Traversal

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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]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

public void levelTraverse(TreeNode root) {
		if (root == null) {
			return;
		}
		LinkedList<TreeNode> queue = new LinkedList<>();
		queue.offer(root);
		while (!queue.isEmpty()) {
			TreeNode node = queue.poll();
			System.out.print(node.val+"  ");
			if (node.left != null) {
				queue.offer(node.left);
			}
			if (node.right != null) {
				queue.offer(node.right);
			}
		}
	}

在来看本题，其实在上述的基础上改造改造就出来了，确实也不怎么难，但是这类题最好熟记，应该把这种当做是个工具类，因为很多题都可以再此基础上改造。确实很多，比如下面的就是下面改造出来的，有好多题是从上面的代码改造出来的。上代码：

    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> lists = new ArrayList<List<Integer>>();
        if (root == null) return lists;
        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        while (!queue.isEmpty()) {
            int len = queue.size();
            List<Integer> list = new ArrayList<Integer>();
            for (int i = 0; i < len; i++) {
                TreeNode node = queue.poll();
                list.add(node.val);
                if (node.left != null) queue.add(node.left);
                if (node.right != null) queue.add(node.right);
            }
            lists.add(list);
        }
        return lists;
    }

LeetCode-Binary Tree Level Order Traversal

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原文地址：http://blog.csdn.net/my_jobs/article/details/47665089