Binary Tree Zigzag Level Order Traversal

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题目：

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]]

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

分析：

层次遍历稍微变了下形状而已

用一个变量记录每层的节点个数就ok了

参考代码：

http://blog.csdn.net/linhuanmars/article/details/24509105

这里LZ用了两个栈！

/** * 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>> res = new ArrayList<List<Integer>>();    if(root==null)        return res;    LinkedList<TreeNode> stack = new LinkedList<TreeNode>();    int level=1;    ArrayList<Integer> item = new ArrayList<Integer>();    item.add(root.val);    res.add(item);    stack.push(root);    while(!stack.isEmpty())    {        LinkedList<TreeNode> newStack = new LinkedList<TreeNode>();        item = new ArrayList<Integer>();        while(!stack.isEmpty())        {            TreeNode node = stack.pop();            if(level%2==0)            {                if(node.left!=null)                {                    newStack.push(node.left);                    item.add(node.left.val);                }                if(node.right!=null)                {                    newStack.push(node.right);                    item.add(node.right.val);                }            }            else            {                if(node.right!=null)                {                    newStack.push(node.right);                    item.add(node.right.val);                }                if(node.left!=null)                {                    newStack.push(node.left);                    item.add(node.left.val);                }                               }        }        level++;        if(item.size()>0)            res.add(item);        stack = newStack;    }    return res;}}

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