Leetcode Binary Tree Zigzag Level Order Traversal

題目：

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]
]

分析：

用兩個棧實現，一個棧pre用來記錄上一層節點的情況，一個棧cur用來記錄本層節點的情況。但是需要注意，奇數層和偶數層的入棧方式不一樣，奇數層的節點彈棧之後先把左節點壓入，偶數層的節點彈棧之後先把右節點壓入。設置一個標誌位來判斷一下即可。

Java代碼實現：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if(root==null)
            return result;
            
        Stack<TreeNode> pre = new Stack<TreeNode>();
        pre.push(root);
        boolean flag = true;
        
        while(!pre.isEmpty())
        {
            ArrayList<Integer> temp = new ArrayList<Integer>();
            Stack<TreeNode> cur = new Stack<TreeNode>();
            while(!pre.isEmpty())
            {
                TreeNode node = pre.pop();
                temp.add(node.val);
                if(flag)
                {
                    if(node.left!=null)
                        cur.push(node.left);
                    if(node.right!=null)
                        cur.push(node.right);
                }
                else
                {
                    if(node.right!=null)
                        cur.push(node.right);
                    if(node.left!=null)
                        cur.push(node.left);
                }
            }
            result.add(temp);
            pre = cur;
            flag = !flag;
        }
        return result;
    }
}