LeetCode-Binary Tree Zigzag Level Order Traversal

Binary Tree Zigzag Level Order Traversal

##题目

####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]
]

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

####OJ’s Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.

Here’s an example:

  1
 / \
2   3
   /
  4
   \
    5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

##解题思路
该题其实还是对树的层次遍历，只不过有一点区别：在奇数层上是从左往右遍历，而在偶数层上是从右往左遍历，因此由于顺序是相反的，我们很容易想到的是用栈来存储每一层的节点。因此这里选用两个栈，一个栈用于读取节点，一个栈用于保存每一层的节点。

##算法代码
代码采用JAVA实现：

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/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
  public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
	    ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<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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