103. Binary Tree Zigzag Level Order Traversal

 

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]
]


与上一题类似
不同在于，奇数层 正着存，偶数层，倒着存。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
import queue
class Solution:
    def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        
        q = queue.Queue()
        if root:
            q.put(root)
        res = []
        level = 0
        while q.qsize():
            sz = q.qsize()
            cur = []
            level+=1
            for i in range(sz):
                top = q.get()
                cur.append(top.val)
               
                    
                if top.left: q.put(top.left)
                if top.right : q.put(top.right)
            if level%2==0:
                res.append(cur[::-1])
            else:
                res.append(cur)
        return res

 

class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        vector<vector<int>> res;
        if(root == nullptr) return res;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()) {
            int cnt = q.size();
            res.push_back(vector<int>());
            for(int i = 0 ; i< cnt; ++i) {
                auto node = q.front(); q.pop();
                if(res.size()%2 != 0) {
                    res.back().push_back(node->val);
                } else {
                    res.back().insert(res.back().begin(),node->val);
                }
                if(node->left!=nullptr) q.push(node->left);
                if(node->right!=nullptr) q.push(node->right);
            }
        }
        return res;
    }
};

 

class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        if(root==null) return res;
        int flag =1;
        queue.add(root);
        while (!queue.isEmpty()) {
            int levlnum = queue.size();
            List<Integer> res_temp = new ArrayList<Integer>();
            for (int i = 0;i<levlnum ;i++ ){ //把每层的左右节点都保存到queue里 
                //并讲当层的值从queue里弹出，加到res_temp 中
            if(queue.peek().left!=null) queue.add(queue.peek().left);
            if(queue.peek().right!=null) queue.add(queue.peek().right);
                if(flag==1)
                res_temp.add(queue.poll().val);
                else
                res_temp.add(0,queue.poll().val);
            }
            res.add(res_temp);
            flag = 1-flag;
        }
        return res;
    }