107. Binary Tree Level Order Traversal II

题目：

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]
]

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

链接： http://leetcode.com/problems/binary-tree-level-order-traversal-ii/

一刷，比102题只多了最后一步，还是用了2个list，第二遍可以试一下DFS

class Solution(object):
    def levelOrderBottom(self, root):
        if not root:
            return []
        result = []
        prev = [root]
        cur = []
        
        while prev:
            for elem in prev:
                if elem.left:
                    cur.append(elem.left)
                if elem.right:
                    cur.append(elem.right)
            result.append([e.val for e in prev])
            prev, cur = cur, []
        result.reverse()
        return result

2/16/2017, Java

错误长的都不知道该怎么写了，而且这段代码效率低得可以不看了

1. List, ArrayList, LinkedList, Queue, Stack的各种区别

2. Interface, Implementing Class的区别，关系

3. List<List<Integer>> ret = new ArrayList<>(); 语法上这是怎么一回事

4. stack(level); level.clear();之后stack里面都是空List，这让用Python的人怎么活。所以Java是怎么存储数据的！

5. 效率提高

public class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {

        Queue<TreeNode> current = new LinkedList<TreeNode>();
        TreeNode p = new TreeNode(-1);
        List<List<Integer>> ret = new ArrayList<>();
        List<Integer> level = new ArrayList<Integer>();
        Stack<List<Integer>> stack = new Stack<>();
        int arraySize = 0;
        
        if (root == null) return ret;

        current.add(root);
        while(current != null && !current.isEmpty()) {
            arraySize = current.size();
            for(int i = 0; i < arraySize; i++) {
                p = current.poll();
                if (p.left != null) {
                    current.add(p.left);
                }
                if (p.right != null) {
                    current.add(p.right);
                }
                level.add(p.val);
            }
            stack.push(level);
            level = new ArrayList<Integer>();
        }
        while(!stack.empty()) {
            ret.add(stack.pop());
        }
        return ret;
    }
}