[LeetCode 103] Binary Tree Zigzag Level Order Traversal

import java.util.ArrayList;
import java.util.List;

/**
 * 
		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]
		]
 *
 */

public class BinaryTreeZigzagLevelOrderTraversal {

	public class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;

		TreeNode(int x) {
			val = x;
		}
	}

//	33 / 33 test cases passed.
//	Status: Accepted
//	Runtime: 236 ms
//	Submitted: 0 minutes ago

	//层次遍历法
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<TreeNode> stack = new ArrayList<TreeNode>();
        List<List<Integer>> zigzags = new ArrayList<List<Integer>>();
        int level = 0;
        if(root == null) {
        	return zigzags;
        }
        stack.add(root);
        while(!stack.isEmpty()) {
        	List<Integer> zigzig = new ArrayList<Integer>();
        	int size = stack.size();
        	level ++;
        	for (int i = 0; i < size; i++) {
				TreeNode node = stack.remove(0);
				if(node.left != null) {
					stack.add(node.left);
				}
				if(node.right != null) {
					stack.add(node.right);
				}
				if(level % 2 == 1) {
					zigzig.add(node.val);
				} else {
					zigzig.add(0, node.val);
				}
			}
        	zigzags.add(zigzig);
        }
        return zigzags;
    }
	public static void main(String[] args) {
		// TODO Auto-generated method stub

	}

}