leetcode -- Binary Tree Zigzag Level Order Traversal

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

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

    3
   / 
  9  20
    /  
   15   7

 return its zigzag level order traversal as:

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

[解题思路]

本题是二叉树层序遍历的变形，二叉树的层序遍历通过使用queue来实现，

一开始我也试图用queue来解这题，第二层很好解决，但到第三层时无法实现从左到右遍历该层

上网搜索了下，发现该题可以使用栈来解决，通过分析执行结果确实是后进先出

这里通过定义leftToRight来表示是从左到右还是从右到左

从左到右：先加left后加right

从右到左：先加right后加left

public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if(root == null){
            return result;            
        }
        Stack<TreeNode> curLvl = new Stack<TreeNode>();
        Stack<TreeNode> nextLvl = new Stack<TreeNode>();
        boolean leftToRight = true;
        curLvl.push(root);
        ArrayList<Integer> output = new ArrayList<Integer>();
        while(!curLvl.empty()){
            TreeNode curNode = curLvl.pop();
            output.add(curNode.val);
            if(leftToRight){
                if(curNode.left != null){
                    nextLvl.add(curNode.left);
                }
                if(curNode.right != null){
                    nextLvl.add(curNode.right);
                }
            } else{
                if(curNode.right != null){
                    nextLvl.add(curNode.right);
                }
                if(curNode.left != null){
                    nextLvl.add(curNode.left);
                }
            }
            if(curLvl.empty()){
                result.add(output);
                output = new ArrayList<Integer>();
                leftToRight = !leftToRight;
                curLvl.addAll(nextLvl);
                nextLvl.clear();
            }
        }
        return result;
    }