【LeetCode】No.103. Binary Tree Zigzag Level Order Traversal -- Java Version

题目链接：https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/

1. 题目介绍（Binary Tree Zigzag Level Order Traversal）

Given the root of a binary tree, return the zigzag level order traversal of its nodes’ values. (i.e., from left to right, then right to left for the next level and alternate between).

【Translate】： 给定二叉树的根，返回其节点值的锯齿级顺序遍历。(例如，从左到右，然后下一个层将从右到左，并在两者之间交替)。

【测试用例】：

示例1：

示例2：

【条件约束】：

2. 题解

2.1 双堆栈

原题解来自于 tjcd 的 JAVA Double Stack Solution.

根据题意，要求Zigzag遍历，而不是像上一个 Binary Tree Level Order Traversal 依次按照顺序将遍历到的点数据存入列表中。而是要满足：①、奇数层的节点数据按照从左到右的顺序存入列表；②、偶数层的节点数据按照从右到左的顺序存入列表。这里就是通过双堆栈进行交替输出，一个堆栈存储奇数层，一个堆栈存储偶数层，关键就在于放入堆栈的顺序，奇序就从左到右放，偶序就从右向左放，最后输出即可。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> ans = new ArrayList<>();
        if(root == null) return ans;
        Stack<TreeNode> s1 = new Stack<>();
        Stack<TreeNode> s2 = new Stack<>();
        s1.push(root);
        
        TreeNode curr = new TreeNode();
        
        while(!s1.isEmpty() || !s2.isEmpty()){
            List<Integer> list = new ArrayList<>();
            
            while(!s1.isEmpty()){
                curr = s1.pop();
                list.add(curr.val);
                if(curr.left != null) s2.push(curr.left);
                if(curr.right != null) s2.push(curr.right);
            }
            ans.add(list);
            list = new ArrayList<>();
            
            while(!s2.isEmpty()){
                curr = s2.pop();
                list.add(curr.val);
                if(curr.right != null) s1.push(curr.right);
                if(curr.left != null) s1.push(curr.left);
            }
            if(!list.isEmpty()) ans.add(list);
        }
        
        return ans;
    }
}
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