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Commit b561507

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feat: update 5 leetcode about binary tree
1 parent 84b6580 commit b561507

5 files changed

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-0
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/*
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* @lc app=leetcode id=222 lang=typescript
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*
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* [222] Count Complete Tree Nodes
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*/
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// @lc code=start
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/**
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* Definition for a binary tree node.
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* class TreeNode {
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* val: number
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* left: TreeNode | null
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* right: TreeNode | null
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* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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* }
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*/
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/**
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* @description: 这里其实对常规二叉树也有效
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* @param {TreeNode} root
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* @return {number}
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*/
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function countNodes(root: TreeNode | null): number {
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if (root === null) return 0;
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const left = countNodes(root.left);
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const right = countNodes(root.right);
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return left + right + 1;
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};
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// @lc code=end
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/*
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* @lc app=leetcode id=230 lang=typescript
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*
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* [230] Kth Smallest Element in a BST
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*/
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// @lc code=start
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/**
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* Definition for a binary tree node.
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* class TreeNode {
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* val: number
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* left: TreeNode | null
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* right: TreeNode | null
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* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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* }
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*/
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/**
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* @description: 中序遍历,迭代的方式
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* 时间复杂度 O(H+k)
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* 空间复杂度 O(H)
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* H 是树的高度
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* @param {TreeNode} root
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* @param {number} k
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* @return {number}
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*/
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function kthSmallest(root: TreeNode | null, k: number): number {
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if (root === null) return null;
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let stack = [root];
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let node = root;
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while (stack.length || node !== null) {
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while (node !== null) {
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stack.push(node);
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node = node.left;
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}
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const cur = stack.pop();
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k--;
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if (k === 0) {
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return cur.val;
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}
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if (cur.right !== null) {
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node = cur.right;
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}
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}
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}
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/**
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* @description: 二叉搜索树中序遍历后输出的值是有序的,利用这一特性
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* 时间复杂度 O(n): 全部遍历一遍
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* 空间复杂度 O(n)? 不确定
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* @param {TreeNode} root
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* @param {number} k
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* @return {number}
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*/
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function kthSmallestByRecursion(root: TreeNode | null, k: number): number {
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let res: number[] = [];
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/**
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* @description: 中序遍历,递归实现
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* @param {TreeNode|null} node
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* @return {void}
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*/
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function inOrderWalker(node: TreeNode | null) {
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if (node === null) return null;
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inOrderWalker(node.left);
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node.val !== null && res.push(node.val);
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inOrderWalker(node.right);
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}
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inOrderWalker(root);
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return res[k - 1];
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};
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// @lc code=end

‎leetcode/268.missing-number.ts

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/*
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* @lc app=leetcode id=268 lang=typescript
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*
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* [268] Missing Number
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*/
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// @lc code=start
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/**
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* @description: 位运算的思路
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* 使用异或
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* 时间复杂度 O(2n+1)
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* 空间复杂度 O(1)
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* @param {number} nums
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* @return {number}
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*/
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function missingNumber(nums:number[]): number {
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const n = nums.length;
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let xor = 0;
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for (let i = 0; i < n; i++) {
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xor ^= nums[i]
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}
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// 注意这里的条件是小于等于,需要计算 n 的情况
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for (let i = 0; i <= n; i++) {
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xor ^= i;
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}
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return xor;
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}
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/**
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* @description: 哈希表的思路
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* 空间换时间
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* 时间复杂度 O(2n)
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* 空间复杂度 O(n)
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* @param {number} nums
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* @return {number}
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*/
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function missingNumber2(nums:number[]): number {
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const arr: number[] = [];
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const n = nums.length;
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for (let i = 0; i < n; i++) {
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// 使用另一个散列表数组来保存所有索引值
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arr[nums[i]] = nums[i];
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}
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for (let i = 0; i < n; i++) {
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// 如果没有,说明缺少这个数字
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if (arr[i] === undefined) return i;
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}
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return n;
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}
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// 时间复杂度应该是 O(n^2)
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// 每次查询会耗费非常多的时间
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function missingNumber3(nums: number[]): number {
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if (nums.length === 1) return nums[0] === 0 ? 1 : 0;
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const n = nums.length;
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// 直接使用连续索引来判断
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for (let i = 0; i < n; i++) {
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if (!nums.includes(i)) return i;
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}
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return n;
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};
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function missingNumber4(nums: number[]): number {
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if (nums.length === 1) return nums[0] === 0 ? 1 : 0;
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const n = nums.length;
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nums.sort((a, b) => a - b);
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// 利用排序后,每一位和对应的索引相等的特点
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for (let i = 0; i < n; i++) {
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if (nums[i] !== i) return i;
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}
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return n;
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};
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function missingNumber5(nums: number[]): number {
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if (nums.length === 1) return nums[0] === 0 ? 1 : 0;
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const n = nums.length;
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nums.sort((a, b) => a - b);
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// 利用排序后,后一位和前一位默认相差是 1 的特点来查询
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for (let i = 1; i < n; i++) {
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if (nums[i] !== nums[i - 1] + 1) return nums[i] - 1;
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}
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return nums[0] === 0 ? n : 0;
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};
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// @lc code=end
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/**
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* Definition for a binary tree node.
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* class TreeNode {
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* val: number
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* left: TreeNode | null
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* right: TreeNode | null
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* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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* }
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*/
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/**
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* @description: 循环的方式,不是递归的方式
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* 时间复杂度 O(n)
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* 空间复杂度 O(1)
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* @param {TreeNode} root
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* @param {number} val
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* @return {TreeNode}
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*/
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function insertIntoBST(root: TreeNode | null, val: number): TreeNode | null {
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if (root === null) return new TreeNode(val);
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let node = root;
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// 循环到尽头位置
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while (node !== null) {
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// 如果节点值比传入值小,就从右边开始找
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if (node.val < val) {
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if (node.right === null) {
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node.right = new TreeNode(val);
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break;
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}
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node = node.right;
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} else {
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if (node.left === null) {
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node.left = new TreeNode(val);
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break;
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}
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node = node.left;
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}
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}
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return root;
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};
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/*
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* @lc app=leetcode id=96 lang=typescript
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*
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* [96] Unique Binary Search Trees
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*/
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// @lc code=start
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/**
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* @description: 数学公式的方式,卡特兰数
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* @param {number} n
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* @return {number}
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*/
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function numTrees(n: number): number {
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let C: number = 1;
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for (let i = 0; i < n; i++) {
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C *= 2 * (2 * i + 1) / (i + 2);
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}
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return C;
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};
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// @lc code=end

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