[LeetCode] 103. Binary Tree Zigzag Level Order Traversal Java

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,null,null,15,7],

    3
   / 
  9  20
    /  
   15   7

return its zigzag level order traversal as:

[
  [3],
  [20,9],
  [15,7]
]

题意及分析：给出一个横向遍历的二叉树，要求给出 按照Z字形横向遍历的每一层 的点。这道题我们可以用stack来求解，简单来讲就是把每一层添加到一个stack里面，然后遍历输出结果的同时将遍历点的子节点添加到一个新的stack里面(先添加左子节点还是右子节点根据上一层添加的顺序)，遍历完之后将新的stack赋值给stack进行下一层的遍历。代码如下：

代码：

/**
 * 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<>();
		 	
		 TreeNode node=root;
		 Stack<TreeNode> stack =new Stack<>();
		 if(node!=null)
			 stack.push(node);
		 int leftOrRight=0;		//0为先遍历左子树，1为右子树
		 while(!stack.isEmpty()){
			 Stack<TreeNode> nextStack =new Stack<>();
			 List<Integer> list=new ArrayList<>();
			 while(!stack.isEmpty()){
				 TreeNode temp=stack.pop();
				 list.add(temp.val);
				 if(leftOrRight==0){
					 if(temp.left!=null)
						 nextStack.push(temp.left);
					 if(temp.right!=null)
						 nextStack.push(temp.right);
				 }else{
					 if(temp.right!=null)
						 nextStack.push(temp.right);
					 if(temp.left!=null)
						 nextStack.push(temp.left);
				 }
			 }
			 if(leftOrRight==0)
				 leftOrRight=1;
			 else {
				leftOrRight=0;
			}
			 res.add(new ArrayList<>(list));
			 //System.out.println(list);
			 list.clear();
			 stack=(Stack<TreeNode>) nextStack.clone ();;
			 nextStack.clear();
		 }
		 return res;
    }
}