lintcode 中等题：binary tree zigzag level order traversal 二叉树的锯齿形层次遍历

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
 
 
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: A list of lists of integer include 
     *          the zigzag level order traversal of its nodes‘ values 
     */
    public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
        // write your code here
        ArrayList<ArrayList<Integer>> tree = new ArrayList<ArrayList<Integer>>();
        if(root == null)
            return tree;
        // rightstack 出栈序列是 左到右该层元素
        Stack<TreeNode> leftstack = new Stack<TreeNode>();
        leftstack.push(root);
        
        // leftstack 出栈序列是 右到左该层元素
        Stack<TreeNode> rightstack = new Stack<TreeNode>();
        boolean left = true;
        while(!leftstack.empty() || !rightstack.empty()){
            ArrayList<Integer> list = new ArrayList<Integer>();
            if(left){
                int size = leftstack.size();
                for(int i=0;i<size;i++){
                    TreeNode node = leftstack.pop();
                    list.add(node.val);
                    if(node.left!=null)
                        rightstack.push(node.left);
                    if(node.right!=null)
                        rightstack.push(node.right);
                    
                }
            }else{
                int size = rightstack.size();
                for(int i=0;i<size;i++){
                    TreeNode node = rightstack.pop();
                    list.add(node.val);
                    if(node.right!=null)
                        leftstack.push(node.right);
                    if(node.left!=null)
                        leftstack.push(node.left);
                    
                }
            }
            left=!left;
            tree.add(list);
        }
        return tree;
    }
}