[leetcode] Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

 

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

https://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/

 

思路：还是层序遍历的思路，用queeu做，插入null分层。返回时Collection.reverse一下。

public class Solution {

    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if (root == null)
            return res;

        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        queue.add(null);

        List<Integer> tmp = new ArrayList<Integer>();
        while (!queue.isEmpty()) {
            TreeNode node = queue.remove();
            if (node == null) {
                res.add(new ArrayList<Integer>(tmp));
                tmp.clear();
                if (!queue.isEmpty())
                    queue.add(null);
            } else {
                // System.out.println(node.val);
                tmp.add(node.val);
                if (node.left != null)
                    queue.add(node.left);
                if (node.right != null)
                    queue.add(node.right);

            }

        }
        Collections.reverse(res);
        return res;

    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(3);
        root.left = new TreeNode(9);
        root.right = new TreeNode(20);
        root.right.left = new TreeNode(15);
        root.right.right = new TreeNode(7);

        System.out.println(new Solution().levelOrderBottom(root));

    }

}

 

参考：

http://blog.csdn.net/xiaozhuaixifu/article/details/12218869