算法（平衡⼆叉树）科普⼆叉树⼆叉树⼆叉数是每个节点最多有两个⼦树,或者是空树(n=0),或者是由⼀个根节点及两个互不相交的,分别称为左⼦树和右⼦树的⼆叉树组成满⼆叉树有两个⾮空⼦树(⼆叉树中的每个结点恰好有两个孩⼦结点切所有叶⼦结点都在同⼀层)也就是⼀个结点要么是叶结点，要么是有两个⼦结点的中间结点。
深度为k且含有2^k-1个结点的⼆叉树完全⼆叉树从左到右依次填充从根结点开始，依次从左到右填充树结点。
除最后⼀层外，每⼀层上的所有节点都有两个⼦节点，最后⼀层都是叶⼦节点。
平衡⼆叉树AVL树[3,1,2,5,9,7]⾸先科普下⼆叉排序树⼜称⼆叉查找树,议程⼆叉搜索树⼆叉排序树的规则⽐本⾝⼤放右边,⽐本⾝⼩放左边平衡⼆叉数⾸先是⼀个⼆叉排序树左右两个⼦树的⾼度差不⼤于1下⾯图中是平衡的情况下⾯是不平衡的情况引⼊公式(LL)右旋function toateRight(AvlNode){let node=AvlNode.left;//保存左节点 AvlNode.left=node.right;node.right=AvlNode;}(RR)左旋function roateLeft(AvlNode){let node=AvlNode.right;//保存右⼦节点AvlNode.right=node.left;node.left=AvlNode;return node;}左右旋⼤图判断⼆叉树是不是平衡树⼆叉树任意结点的左右⼦树的深度不超过1深度计算定义⼀个初始化的⼆叉树var nodes = {node: 6,left: {node: 5,left: {node: 4},right: {node: 3}},right: {node: 2,right: {node: 1}}}//计算⾼度const treeDepth = (root) => {if (root == null) {return 0;}let left = treeDepth(root.left)let right = treeDepth(root.right)return 1+(left>right?left:right)}//判断深度const isTree=(root)=>{if (root == null) {return true;}let left=treeDepth(root.left)let right=treeDepth(root.right)let diff=left-right;if (diff > 1 || diff < -1) {return false}return isTree(root.left)&&isTree(root.right) }console.log(isTree(nodes))判断⼆叉数是不是搜索⼆叉树//第⼀种 //中序遍历let last=-Infinity;const isValidBST=(root)=>{if (root == null) {return true;}//先从左节点开始if (isValidBST(root.left)) {if (last < root.node) {last=root.node;return isValidBST(root.right)}}return false}console.log(isValidBST(nodes))//第⼆种const isValidBST = root => {if (root == null) {return true}return dfs(root, -Infinity, Infinity)}const dfs = (root, min, max) => {if (root == null) {return true}if (root.node <= min || root.node >= max) {return false}return dfs(root.left, min, root.node) && dfs(root.right, root.node, max)}console.log(isValidBST(nodes))实现⼀个⼆叉树实现了⼆叉树的添加,删除,查找,排序//⼆叉树结点class TreeNode {constructor(n, left, right){this.n = n;this.left = left;this.right = right;}}//⼆叉树class BinaryTree {constructor(){this.length = 0;this.root = null;this.arr = [];}//添加对外⼊⼝，⾸个参数是数组，要求数组⾥都是数字，如果有不是数字则试图转成数字，如果有任何⼀个⽆法强制转成数字，则本操作⽆效 addNode(){let arr = arguments[0];if(arr.length == 0) return false;return this.judgeData('_addNode', arr)}//删除结点deleteNode(){let arr = arguments[0];if(arr.length == 0) return false;return this.judgeData('_deleteNode', arr)}//传值判断，如果全部正确，则全部加⼊叉树judgeData(func, arr){let flag = false;//任何⼀个⽆法转成数字，都会失败if(arr.every(n => !Number.isNaN(n))){let _this = this;arr.map(n => _this[func](n));flag = true;}return flag;}//添加的真实实现_addNode(n){n = Number(n);let current = this.root;let treeNode = new TreeNode(n, null, null);if(this.root === null){this.root = treeNode;}else {current = this.root;while(current){let parent = current;if(n < current.n){current = current.left;if(current === null){parent.left = treeNode;}}else {current = current.right;if(current === null){parent.right = treeNode;}}}}this.length++;return treeNode;}//删除节点的真实实现_deleteNode(n){n = Number(n);if(this.root === null){return;}//查找该节点，删除节点操作⽐较复杂，为排除找不到被删除的节点的情况，简化代码，先保证该节点是存在的，虽然这样做其实重复了⼀次查询了，但⼆叉树的查找效率很⾼，这是可接受的let deleteNode = this.findNode(n);if(!deleteNode){return;}//如果删除的是根节点if(deleteNode === this.root){if(this.root.left === null && this.root.right === null){this.root = null;}else if(this.root.left === null){this.root = this.root.right;}else if(this.root.right === null){this.root = this.root.left;}else {let [replaceNode, replacePNode, rp] = this.findLeftTreeMax(deleteNode);replacePNode[rp] = null;replaceNode.left = this.root.left;replaceNode.right = this.root.right;this.root = replaceNode;}}else {//被删除的⽗节点，⼦节点在⽗节点的位置p，有left,right两种可能let [deleteParent, p] = this.findParentNode(deleteNode);if(deleteNode.left === null && deleteNode.right === null){deleteParent[p] = null;}else if(deleteNode.left === null){deleteParent[p] = deleteNode.right;}else if(deleteNode.right === null){deleteParent[p] = deleteNode.left;}else {//⽤来替换被删除的节点，⽗节点，节点在⽗节点的位置let [replaceNode, replacePNode, rp] = this.findLeftTreeMax(deleteNode);if(replacePNode === deleteNode){deleteParent[p] = replaceNode;}else {deleteParent[p] = replaceNode;replacePNode.right = null;}replacePNode[rp] = null;replaceNode.left = deleteNode.left;replaceNode.right = deleteNode.right;}}this.length--;}//查找findNode(n){let result = null;let current = this.root;while(current){if(n === current.n){result = current;break;}else if(n < current.n){current = current.left;}else {current = current.right;}}return result;}//查找⽗节点findParentNode(node){let [parent, child, p] = [null, null, null];if(this.root !== node){parent = this.root;if(node.n < parent.n){child = parent.left;p = 'left';}else {child = parent.right;p = 'right';}while(child){if(node.n === child.n){break;}else if(node.n < child.n){parent = child;child = parent.left;p = 'left';}else {parent = child;child = parent.right;p = 'right';}}}return [parent, p];}//查找当前有左⼦树的节点的最⼤值的节点M，如有A个节点被删除，M是最接近A点之⼀（还有⼀个是右⼦树节点的最⼩值） findLeftTreeMax(topNode){let [node, parent, p] = [null, null, null];if(this.root === null || topNode.left === null){return [node, parent, p];}parent = topNode;node = topNode.left;p = 'left';while(node.right){parent = node;node = node.right;p = 'right';}return [node, parent, p];}//查找最⼤值maxValue(){if(this.root !== null){return this._findLimit('right');}}//查找最⼩值minValue(){if(this.root !== null){return this._findLimit('left');}}//实现查找特殊值_findLimit(pro){let n = this.root.n;let current = this.root;while(current[pro]){current = current[pro];n = current.n;}return n;}//中序排序,并⽤数组的形式显⽰sortMiddleToArr(){this._sortMiddleToArr(this.root);return this.arr;}//中序⽅法_sortMiddleToArr(node){if(node !== null){this._sortMiddleToArr(node.left);this.arr.push(node.n);this._sortMiddleToArr(node.right);}}//打印⼆叉树对象printNode(){console.log(JSON.parse(JSON.stringify(this.root)));}}//测试var binaryTree = new BinaryTree();binaryTree.addNode([50, 24, 18, 65, 4, 80, 75, 20, 37, 40, 60]);binaryTree.printNode();//{n: 50, left: {…}, right: {…}}console.log(binaryTree.maxValue());//80console.log(binaryTree.minValue());//4console.log(binaryTree.sortMiddleToArr());// [4, 18, 20, 24, 37, 40, 50, 60, 65, 75, 80] binaryTree.deleteNode([50]);binaryTree.printNode();//{n: 40, left: {…}, right: {…}}排序复习function ArrayList() {this.array = [];}ArrayList.prototype = {constructor: ArrayList,insert: function(item) {this.array.push(item);},toString: function() {return this.array.join();},swap: function(index1, index2) {var aux = this.array[index2];this.array[index2] = this.array[index1];this.array[index1] = aux;},//冒泡排序bubbleSort: function() {var length = this.array.length;for (var i = 0; i < length; i++) {for (var j = 0; j < length - 1 - i; j++) {if (this.array[j] > this.array[j + 1]) {this.swap(j, j + 1);}}}},//选择排序selectionSort: function() {var length = this.array.length;var indexMin;for (var i = 0; i < length - 1; i++) {indexMin = i;for (var j = i; j < length; j++) {if (this.array[indexMin] > this.array[j]) {indexMin = j;}}if (indexMin !== i) {this.swap(indexMin, i);}}},//插⼊排序insertionSort: function() {var length = this.array.length;var j;var temp;for (var i = 1; i < length; i++) {temp = this.array[i];j = i;while (j > 0 && this.array[j - 1] > temp) {this.array[j] = this.array[j - 1];j--;}this.array[j] = temp;}},//归并排序mergeSort: function() {function mergeSortRec(array) {var length = array.length;if (length === 1) {return array;}var mid = Math.floor(length / 2);var left = array.slice(0, mid);var right = array.slice(mid, length);return merge(mergeSortRec(left), mergeSortRec(right)); }function merge(left, right) {var result = [];var il = 0;var ir = 0;while (il < left.length && ir < right.length) {if (left[il] < right[ir]) {result.push(left[il++]);} else {result.push(right[ir++]);}}while (il < left.length) {result.push(left[il++]);}while (ir < right.length) {result.push(right[ir++]);}return result;}this.array = mergeSortRec(this.array);},//快速排序quickSort:function(){function sort(array){if (array.length <= 1) {return array;}var pivotIndex = Math.floor(array.length/2);var pivot = array.splice(pivotIndex,1)[0];var left = [];var right = [];for(var i = 0; i < array.length; i++){if (array[i] < pivot) {left.push(array[i]);}else{right.push(array[i]);}}return sort(left).concat([pivot],sort(right));}this.array = sort(this.array);}};...................................................................................................................############################################################################ ###################################################################################。
