Leetcode:BinaryTree level order Traversal and Zigzag 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

return its bottom-up level order traversal as:

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

ZigZag traversal  就是 第一行 from left to right,the second row is from right to left

return its bottom-up level order traversal as:

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

把这个写到博客是其中有2个处理技巧 非常重要 I never think of these ways to deal with these problems

the first problem: When the queue size changes every time,but we only need the original size how to deal it

the second problem:how to deal with the zigzag traversal(见下面的解释)

int size = queue.size;for(int i=0;i<size;i++);

while(!queue.isEmpty()) {            ArrayList<Integer> list = new ArrayList<Integer>();            TreeNode node = queue.poll();            for (int i = 0; i < queue.size(); i++) {//如果 node的左孩子和右孩子任意一个不为空的话，queue的size在这里一直            //在变,但是第一次刷这一题的时候 也遇到这个问题，第二次刷的时候，还是遇到这样的问题，所以for循环条件需要改变                list.add(node.val);                if (node.left!= null) {                    if (node.left != null) queue.offer(node.left);                }                if(node.right!=null){                    if(node.right!=null) queue.offer(node.right);                }            }            result.add(list);        }

zigzag遍历的话  就需要我们设置一个flag 每次循环遍历一个链表之后改变一下flag的值 这段代码可以加在result.add(list）上面 刚开始flag设为true

if(!flag){         Collections.reverse(list);         flag = true;         }else{               flag = false;               }