【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]
]
运用栈的先进先出特点，并保证每层元素进栈的顺序是从左向右，或者从右向左

public class Solution {
    public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
        ArrayList<Integer> ai = new ArrayList<Integer>();
        ArrayList<ArrayList<Integer>> re = new ArrayList<ArrayList<Integer>>();
        if(root==null)
            return re;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        Stack<TreeNode> tempStack = new Stack<TreeNode>();
        stack.push(root);
        int i=0;
        while(!stack.isEmpty()){
            TreeNode tempnode = stack.peek();
            stack.pop();
            ai.add(tempnode.val);
            //根据层序的层数来确定是从左向右添加还是从右向左添加
            //保证stack中的顺序是连续的
            if(i%2==0){
                if(tempnode.left!=null){
                    tempStack.push(tempnode.left);
                }
                if(tempnode.right!=null){
                    tempStack.push(tempnode.right);
                }
            }else{
                if(tempnode.right!=null){
                    tempStack.push(tempnode.right);
                }
                if(tempnode.left!=null){
                    tempStack.push(tempnode.left);
                }
            }
            
            if(stack.isEmpty()){
                i++;
                stack=tempStack;
                
                tempStack = new Stack<TreeNode>();
                re.add(ai);
                ai=new ArrayList<Integer>();
            }
            
        }
        return re;
        
    }
}