init commit

Author: yuanyu90221

.github/workflows/java.yml

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+name: Java
+
+on:
+  push:
+    branches:
+      - master
+  pull_request:
+    branches:
+      - master
+
+jobs:
+  build:
+    runs-on: ubuntu-latest
+    steps:
+      - name: Checkout
+        uses: actions/checkout@v3
+      - name: Setup JDK 11
+        uses: actions/setup-java@v3
+        with:
+          distribution: 'adopt'
+          java-version: '11'
+      - name: Grant execution permission for gradlew
+        run: chmod +x gradlew
+      - name: Cache Gradle packages
+        uses: actions/cache@v3
+        with:
+          path: ~/.gradle/caches
+          key: ${{ runner.os }}-gradle-${{ github.event.repository.name }}-${{ hashFiles('**/*.gradle')}}
+          restore-keys: ${{ runner.os }}-gradle-${{ github.event.repository.name }}
+      - name: Build with Gradlew
+        run: ./gradlew build
+      - name: Test with Gradlew
+        run: ./gradlew :test --tests "SolutionTest"

.gitignore

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+.gradle
+build/
+!gradle/wrapper/gradle-wrapper.jar
+!**/src/main/**/build/
+!**/src/test/**/build/
+
+### IntelliJ IDEA ###
+.idea/modules.xml
+.idea/jarRepositories.xml
+.idea/compiler.xml
+.idea/libraries/
+*.iws
+*.iml
+*.ipr
+out/
+!**/src/main/**/out/
+!**/src/test/**/out/
+
+### Eclipse ###
+.apt_generated
+.classpath
+.factorypath
+.project
+.settings
+.springBeans
+.sts4-cache
+bin/
+!**/src/main/**/bin/
+!**/src/test/**/bin/
+
+### NetBeans ###
+/nbproject/private/
+/nbbuild/
+/dist/
+/nbdist/
+/.nb-gradle/
+
+### VS Code ###
+.vscode/
+
+### Mac OS ###
+.DS_Store

.husky/pre-commit

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+#!/bin/bash
+./gradlew :test --tests "SolutionTest"

.idea/gradle.xml

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+# java_valid_BST
+
+Given the `root` of a binary tree, *determine if it is a valid binary search tree (BST)*.
+
+A **valid BST** is defined as follows:
+
+- The left subtree of a node contains only nodes with keys **less than** the node's key.
+- The right subtree of a node contains only nodes with keys **greater than** the node's key.
+- Both the left and right subtrees must also be binary search trees.
+
+## Examples
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/12/01/tree1.jpg](https://assets.leetcode.com/uploads/2020/12/01/tree1.jpg)
+
+```
+Input: root = [2,1,3]
+Output: true
+
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/12/01/tree2.jpg](https://assets.leetcode.com/uploads/2020/12/01/tree2.jpg)
+
+```
+Input: root = [5,1,4,null,null,3,6]
+Output: false
+Explanation: The root node's value is 5 but its right child's value is 4.
+
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[1, 104]`.
+- `231 <= Node.val <= 231 - 1`
+
+## 解析
+
+題目給定一個二元樹的根結點 root
+
+想要判斷這個由 root 所形成的樹是不是一棵二元搜尋樹
+
+如果 root 所形成的樹是一個二元搜尋樹有以下特性
+
+1. root.Left 所形成的樹內所有結點值都小於 root.Val
+2. root.Right 所形成的樹內所有結點值都大於 root.Val
+3. root.Left 與 root.Right 所形成的樹都是二元搜尋樹
+
+最直覺的想法
+
+每次遇到一個結點
+
+檢查這個結點的左子樹有沒有都小於這個 root 結點
+
+檢查這個結點的右子樹有沒有都大於這個 root 結點 
+
+這樣每次都要走訪 n個 所以時間複雜度 是 O($n^2$)
+
+如果不想要每次都要走訪所有走過的結點則必須思考以下特性
+
+對一個二元搜尋樹的某個結點 node
+
+node.left 所形成樹的上界是 node.Val ，因為所有 node.Left 所形成結點都必須小於 node.Val
+
+所以對於 node.Left 所形成樹 只要檢查是否都有小於 node.Val 即可
+
+同樣的，node.Right 所形成樹的下界是 node.Val ，因為所有 node.Right 所形成結點都必須大於 node.Val
+
+所以對於 node.Right 所形成樹 只要檢查是否都有大於 node.Val 即可
+
+![](https://i.imgur.com/gM2lJ9d.png)
+
+一開始的根結點 沒有上下界相當於 上界是 infinity , 下界是 -infinity
+
+每次只要檢查目前結點有沒有介於上下界
+
+每次往左結點走更新上界是目前結點
+
+每次往右結點走更新下界是目前結點
+
+## 程式碼
+
+```java
+class Solution {
+  /**
+   * Definition for a binary tree node.
+   * public class TreeNode {
+   *     int val;
+   *     TreeNode left;
+   *     TreeNode right;
+   *     TreeNode() {}
+   *     TreeNode(int val) { this.val = val; }
+   *     TreeNode(int val, TreeNode left, TreeNode right) {
+   *         this.val = val;
+   *         this.left = left;
+   *         this.right = right;
+   *     }
+   * }
+   */
+  public boolean isValidBST(TreeNode root) {
+    if (root == null) {
+      return true;
+    }
+    return checkBST(root, Double.MAX_VALUE, -Double.MAX_VALUE);
+  }
+  public boolean checkBST(TreeNode root, double upperbound, double lowerbound) {
+    if (root == null) {
+      return true;
+    }
+    double cur = (double)(root.val);
+    if (cur >= upperbound || cur <= lowerbound) {
+      return false;
+    }
+    return checkBST(root.left, cur, lowerbound) && checkBST(root.right, upperbound, cur);
+  }
+}
+```
+
+## 困難點
+
+1. 理解二元搜尋樹的定義
+2. 透過理解二元搜尋樹的定義來減少檢查的範圍
+
+## Solve Point
+
+- [x]  Understand what problem need to be solve
+- [x]  Analysis Complexity