甜甜马里奥赛事总结第一天——Leetcode Weekly Contest 169 3562 / 5932

目前比赛停滞在一半左右的名词，每次只是两道题

但这次虽然也打了两道题却感觉吃力，最近荒废刷题的原因，真的是一日不刷自己知道，三日不刷，比赛知道

可以不打比赛，但是不可以不写题解。只有思考才能进步

题解正文：

第二题：

知识点：二叉搜索树是有序树，左侧节点比跟节点小，右侧大，这道排序题有天然优势

问题：如何获得二叉树的值，遍历总是有问题，因为比赛的时候我的遍历代码是试出来的，不知道为啥能成功

原因：

值传递 VS 引用传递：

区分不清导致

值传递只要变现是在内存中新开辟一块变量，给赋值操作的对象，更改新对象不会影响旧对象

引用传递对新值的更改会连同旧的一起，所以整个递归操作一直是对同一个list进行添加，因此最后直接输出即可不用再次赋值

改进点：或许可以同步遍历两颗树，遍历完排序完，而不是先存储后排序

改进点：

5296. All Elements in Two Binary Search Trees

Given two binary search trees root1 and root2.

Return a list containing all the integers from both trees sorted in ascending order.

Example 1:

Input: root1 = [2,1,4], root2 = [1,0,3]
Output: [0,1,1,2,3,4]

Example 2:

Input: root1 = [0,-10,10], root2 = [5,1,7,0,2]
Output: [-10,0,0,1,2,5,7,10]

Example 3:

Input: root1 = [], root2 = [5,1,7,0,2]
Output: [0,1,2,5,7]

Example 4:

Input: root1 = [0,-10,10], root2 = []
Output: [-10,0,10]

Example 5:

Input: root1 = [1,null,8], root2 = [8,1]
Output: [1,1,8,8]

Constraints:

Each tree has at most 5000 nodes.
Each node's value is between [-10^5, 10^5].

赛事代码：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def getAllElements(self, root1, root2):
        """
        :type root1: TreeNode
        :type root2: TreeNode
        :rtype: List[int]
        """
        list1 = []
        list2 = []
        if root2==None:
            return self.getVal(root1, list1)
        if root1==None:
            return self.getVal(root2, list2)
        list1 = self.getVal(root1, list1)
        list2 = self.getVal(root2, list2)
        print(list1)
        print(list2)
        len_1 = len(list1)
        len_2 = len(list2)
        i, j = 0, 0
        ans = []
        while i != len_1 and j != len_2:
            if list1[i] <= list2[j]:
                ans.append(list1[i])
                i += 1
            else:
                ans.append(list2[j])
                j += 1
        if i != len_1:
            ans += list1[i:]
        else:
            ans += list2[j:]
        return ans

        print(list2)

    def getVal(self, root, res):
        if root.left:
            root3 = root.left
            res = self.getVal(root3, res)
        res.append(root.val)
        if root.right:
            root4 = root.right
            res = self.getVal(root4, res)
        return res

第三题：

5297. Jump Game III

Given an array of non-negative integers arr, you are initially positioned at start index of the array. When you are at index i, you can jump to i + arr[i] or i - arr[i], check if you can reach to any index with value 0.

Notice that you can not jump outside of the array at any time.

Example 1:

Input: arr = [4,2,3,0,3,1,2], start = 5
Output: true
Explanation: 
All possible ways to reach at index 3 with value 0 are: 
index 5 -> index 4 -> index 1 -> index 3 
index 5 -> index 6 -> index 4 -> index 1 -> index 3

Example 2:

Input: arr = [4,2,3,0,3,1,2], start = 0
Output: true 
Explanation: 
One possible way to reach at index 3 with value 0 is: 
index 0 -> index 4 -> index 1 -> index 3

Example 3:

Input: arr = [3,0,2,1,2], start = 2
Output: false
Explanation: There is no way to reach at index 1 with value 0.

Constraints:

1 <= arr.length <= 5 * 10^4
0 <= arr[i] < arr.length
0 <= start < arr.length