Lunski's Clutter

終於來到我目前施行的課表了，一般來說是一週推拉腿推拉腿，週日休息的節奏，但考量到週三可能要復健，一五有時會被貼紮或拉筋，而我最需加強腿部。

從摸索跟不斷追尋下，我理解到課表需要修正，由於週三一直都是去復健，而去健身房時間大約只能做
3個動作，我規劃了一天上肢/下肢各兩個動作，搭配一個胸腹，這份課表目的是變成上下分化訓練，在某部分痠痛時訓練另一半。

Given the root of a binary tree, return the inorder traversal of its nodes’ values.

Example 1:

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Input: root = [1,null,2,3]
Output: [1,3,2]

Example 2:

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Input: root = []
Output: []

Example 3:

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

Example 4:

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Input: root = [1,2]
Output: [2,1]

Example 5:

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Input: root = [1,null,2]
Output: [1,2]

在吃飯聽到這首歌，很喜歡主唱細膩的聲線，MV在說男女遇到災難時的反應，但我想是在說更普羅大眾的-當感情遇到考驗時你還會選擇繼續相信對方嗎？

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

push(x) – Push element x onto stack.
pop() – Removes the element on top of the stack.
top() – Get the top element.
getMin() – Retrieve the minimum element in the stack.

Example 1:

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Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

Output
[null,null,null,null,-3,null,0,-2]

Explanation

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MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top();    // return 0
minStack.getMin(); // return -2