[LeetCode]Binary Tree Zigzag Level Order Traversal

Question
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]
]

本题难度Medium。迭代法用队列实现，递归法可以用栈和队列分别实现。

1、迭代+队列

【复杂度】
时间 O(b^(h+1)-1) 空间 O(b^h)

【思路】
与​​[LeetCode]Binary Tree Level Order Traversal​​差不多，区别点就在于向​​list​​​加入​​n.val​​的方式：

1. 如果层序​​level​​为偶数，逆序加入
2. 如果层序​​level​​为奇数，顺序加入

【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) {
        //require
        List<List<Integer>> ans=new LinkedList<>();
        Queue<TreeNode> q=new LinkedList<>();
        if(root!=null)q.add(root);
        int level=1;
        //invariant
        while(!q.isEmpty()){
            List<Integer> list=new LinkedList<>();
            int size=q.size();
            for(int i=0;i<size;i++){
                TreeNode n=q.remove();
                if(level%2==0)
                    list.add(0,n.val);
                else
                    list.add(n.val);
                if(n.left!=null)q.add(n.left);
                if(n.right!=null)q.add(n.right);
            }
            ans.add(list);
            level++;
        }
        //ensure
        return ans;
    }
}

2、栈+递归

【复杂度】
时间 O(b^(h+1)-1) 空间 O(b^h)

【思路】
根据层序进行区分：

1. 如果层序​​level​​为奇数，节点的儿子先左儿子后右儿子入栈
2. 如果层序​​level​​为偶数，节点的儿子先右儿子后左儿子入栈

【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) {
        //require
        List<List<Integer>> ans=new LinkedList<>();
        Stack<TreeNode> stack=new Stack<>();
        if(root!=null)stack.push(root);
        //invariant
        helper(stack,1,ans);
        //ensure
        return ans;
    }
    public void helper(Stack<TreeNode> stack,int level,List<List<Integer>> ans){
        //base case
        if(stack.isEmpty())
            return;

        Stack<TreeNode> nextStack=new Stack<>();
        List<Integer> list=new LinkedList<>();
        while(!stack.isEmpty()){
           TreeNode n=stack.pop();
           list.add(n.val);
           if(level%2==0){
               if(n.right!=null)nextStack.push(n.right);
               if(n.left!=null)nextStack.push(n.left);
           }else{
               if(n.left!=null)nextStack.push(n.left);
               if(n.right!=null)nextStack.push(n.right);
           }
        }
        ans.add(list);
        helper(nextStack,level+1,ans);
    }
}