LeetCode-102&107.Binary Tree Level Order Traversal

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

102：

return its level order traversal as:

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

107：

return its bottom-up level order traversal as:

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

这道题与LeetCode 104的解法(http://blog.csdn.net/zmq570235977/article/details/51210982)基本一样

102：

public IList<IList<int>> LevelOrderBottom(TreeNode root)        {            IList<IList<int>> res = new List<IList<int>>();            if (root == null)                return res;            int num;            TreeNode node;            Queue<TreeNode> q = new Queue<TreeNode>();            q.Enqueue(root);            while ((num = q.Count) > 0)            {                IList<int> list = new List<int>();                for (int i = 0; i < num; i++)                {                    node = q.Dequeue();                    if (node.left != null)                        q.Enqueue(node.left);                    if (node.right != null)                        q.Enqueue(node.right);                    list.Add(node.val);                }                res.Add(list);            }            return res;        }

107：将

res.Add(list);

换成

res.Insert(0, list);

即可。