1. LeetCode 103

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).**Example 1:**

Input:root = [3,9,20,null,null,15,7]Output:[[3],[20,9],[15,7]]

**Example 2:**

Input:root = [1]Output:[[1]]

**Example 3:**

Input:root = []Output:[]

**Constraints:**

- The number of nodes in the tree is in the range [0, 2000].
- -100 <= Node.val <= 100

2. LeetCode 134

There are n gas stations along a circular route, where the amount of gas at the ith station is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from the ith station to its next (i + 1)th station. You begin the journey with an empty tank at one of the gas stations.

Given two integer arrays gas and cost, return

*the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return*-1. If there exists a solution, it is**guaranteed**to be**unique**

**Example 1:**

Input:gas = [1,2,3,4,5], cost = [3,4,5,1,2]Output:3Explanation:Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4 Travel to station 4. Your tank = 4 - 1 + 5 = 8 Travel to station 0. Your tank = 8 - 2 + 1 = 7 Travel to station 1. Your tank = 7 - 3 + 2 = 6 Travel to station 2. Your tank = 6 - 4 + 3 = 5 Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3. Therefore, return 3 as the starting index.

**Example 2:**

Input:gas = [2,3,4], cost = [3,4,3]Output:-1Explanation:You can't start at station 0 or 1, as there is not enough gas to travel to the next station. Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4 Travel to station 0. Your tank = 4 - 3 + 2 = 3 Travel to station 1. Your tank = 3 - 3 + 3 = 3 You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3. Therefore, you can't travel around the circuit once no matter where you start.

**Constraints:**

- n == gas.length == cost.length
- 1 <= n <= 105
- 0 <= gas[i], cost[i] <= 104