題目：
請實現(xiàn)一個函數(shù)按照之字形打印二叉樹，即第一行按照從左到右的順序打印，第二層按照從右至左的順序打印，第三行按照從左到右的順序打印，其他行以此類推。

思路：
層次遍歷通過一個層數(shù)來控制左-右或者右-左,右到左相當(dāng)于一個reverse所以可以根據(jù)棧的特性來完成這道題.

代碼：

/**
 * Created by Hammy on 2018/2/1.
 */
public class Print
{
    public ArrayList<ArrayList<Integer> > Print(TreeNode pRoot) {
        ArrayList<ArrayList<Integer>> arrayLists = new ArrayList<>();
        if(pRoot == null)
            return arrayLists;
        //代表當(dāng)前層數(shù)
        int postNumber = 1 ;
        //放奇數(shù)層的棧
        Stack<TreeNode> stack1 = new Stack<>();
        //偶數(shù)層
        Stack<TreeNode> stack2 = new Stack<>();
        stack1.push(pRoot);

        while(!stack1.isEmpty() || !stack2.isEmpty()){
            //代表奇數(shù)
            if(postNumber % 2 != 0){
                ArrayList<Integer> arrayList = new ArrayList<>();
                while(!stack1.isEmpty()){
                    TreeNode node = stack1.pop();
                    if(node != null){
                        arrayList.add(node.val);
                        stack2.push(node.left);
                        stack2.push(node.right);
                    }
                }
                if(!arrayList.isEmpty()){
                    arrayLists.add(arrayList);
                    postNumber++;
                }
            }else{
                ArrayList<Integer> arrayList = new ArrayList<>();
                while(!stack2.isEmpty()){
                    TreeNode node = stack2.pop();
                    if(node != null){
                        arrayList.add(node.val);
                        stack1.push(node.right);
                        stack1.push(node.left);
                    }
                }
                if(!arrayList.isEmpty()){
                    arrayLists.add(arrayList);
                    postNumber++;
                }
            }
        }

        return arrayLists;
    }
    }