add: 验证二叉搜索树

README.md

@@ -281,6 +281,11 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
   - LeetCode 199. 二叉树的右视图 https://leetcode-cn.com/problems/binary-tree-right-side-view/
   - LintCode 760. 二叉树的右视图 https://www.lintcode.com/problem/binary-tree-right-side-view/description
 
+- [验证二叉搜索树](src/tree/validate-binary-search-tree.js)
+
+  - LeetCode 98. 验证二叉搜索树 https://leetcode-cn.com/problems/validate-binary-search-tree/
+  - LintCode 95. 验证二叉查找树 https://www.lintcode.com/problem/validate-binary-search-tree/
+
 ## 链表
 
 - [合并K个排序链表](src/list/merge-k-sorted-lists.js)

src/tree/validate-binary-search-tree.js

@@ -0,0 +1,16 @@
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val) {
+ *     this.val = val;
+ *     this.left = this.right = null;
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {boolean}
+ */
+export const isValidBST = function (root, lower = -Infinity, upper = Infinity) {
+  if (root === null) return true
+  if (root.val <= lower || root.val >= upper) return false
+  return isValidBST(root.left, lower, root.val) && isValidBST(root.right, root.val, upper)
+}

test/tree/Tree.js

@@ -0,0 +1,40 @@
+import TreeNode from './TreeNode'
+
+export default class Tree {
+  constructor () {
+    this.root = null
+    this.queue = []
+    this.insertNum = 0
+  }
+
+  insert (val) {
+    this.insertNum++ // 插入次数加1
+    const node = (!val && typeof val === 'object') ? null : new TreeNode(val) // 判断是否为空节点
+
+    if (!this.root) { // 判断根节点是否存在
+      this.root = node // 插入根节点
+      node && this.queue.push(this.root) // 非空节点入列
+    } else { // 插入非根节点
+      const parent = this.queue[0] // 被插入的父节点
+      if (!(this.insertNum % 2)) { // 通过插入次数判断左右
+        parent.left = node // 插入左边
+        node && this.queue.push(parent.left) // 非空节点入列
+      } else {
+        parent.right = node // 插入右边
+        node && this.queue.push(parent.right) // 非空节点入列
+        this.queue.shift() // 当前父节点parent 已经不可能再插入子节点，故出列
+      }
+    }
+    return this
+  }
+
+  static arrToTree (arr) {
+    const tree = new Tree()
+
+    for (const n in arr) {
+      tree.insert(arr[n])
+    }
+
+    return tree.root
+  }
+}

test/tree/TreeNode.js

@@ -0,0 +1,7 @@
+export default class TreeNode {
+  constructor (val) {
+    this.val = val
+    this.left = null
+    this.right = null
+  }
+}

test/tree/binary-tree-right-side-view.test.js

@@ -1,48 +1,8 @@
 import { rightSideView } from '../../src/tree/binary-tree-right-side-view'
+import Tree from './Tree'
 
 test('二叉树的右视图', () => {
-  function TreeNode (val) {
-    this.val = val
-    this.left = null
-    this.right = null
-  }
-
-  class Tree {
-    constructor () {
-      this.root = null
-      this.queue = []
-      this.insertNum = 0
-    }
-
-    insert (val) {
-      this.insertNum++ // 插入次数加1
-      const node = (!val && typeof val === 'object') ? null : new TreeNode(val) // 判断是否为空节点
-
-      if (!this.root) { // 判断根节点是否存在
-        this.root = node // 插入根节点
-        node && this.queue.push(this.root) // 非空节点入列
-      } else { // 插入非根节点
-        const parent = this.queue[0] // 被插入的父节点
-        if (!(this.insertNum % 2)) { // 通过插入次数判断左右
-          parent.left = node // 插入左边
-          node && this.queue.push(parent.left) // 非空节点入列
-        } else {
-          parent.right = node // 插入右边
-          node && this.queue.push(parent.right) // 非空节点入列
-          this.queue.shift() // 当前父节点parent 已经不可能再插入子节点，故出列
-        }
-      }
-      return this
-    }
-  }
-
   const arr = [1, 2, 3, null, 5, null, 4]
 
-  const tree = new Tree()
+  expect(rightSideView(Tree.arrToTree(arr))).toEqual([1, 3, 4])
-
-  for (const n in arr) {
-    tree.insert(arr[n])
-  }
-
-  expect(rightSideView(tree.root)).toEqual([1, 3, 4])
 })

test/tree/validate-binary-search-tree.test.js

@@ -0,0 +1,12 @@
+import { isValidBST } from '../../src/tree/validate-binary-search-tree'
+import Tree from './Tree'
+
+test('验证二叉搜索树', () => {
+  expect(isValidBST(Tree.arrToTree([1, 1]))).toBe(false)
+  expect(isValidBST(Tree.arrToTree([2, 1, 3]))).toBe(true)
+  expect(isValidBST(Tree.arrToTree([5, 1, 4, null, null, 3, 6]))).toBe(false)
+  expect(isValidBST(Tree.arrToTree([-1]))).toBe(true)
+  expect(isValidBST(Tree.arrToTree([2, 1, 4, null, null, 3, 5]))).toBe(true)
+  expect(isValidBST(null)).toBe(true)
+  expect(isValidBST(Tree.arrToTree([10, 5, 15, null, null, 6, 20]))).toBe(false)
+})