概述
樹在前端的重要性就不言而喻了,隨處可見,vdom,dom tree,render tree,有時候前後端互動中也會收到具有遞迴性質的tree結構資料,需要注意一點的是es6中雖然出現了set和map資料結構,但其實現和其它語言(例如java中)底層實現不同,在chrome的 v8中其實現基於hash,利用空間換時間的思想,畢竟查詢起來hash O(1)而紅黑樹O(LgN)。但是紅黑樹作為一種經典且重要的資料結構,綜合優勢比較好,curd操作以及空間消耗在大量資料下優勢就體現出來了。
js實現
const RED = true;
const BLACK = false;
class Node {
constructor(key, value) {
this.key = key;
this.value = value;
this.left = null;
this.right = null;
this.color = RED;
}
}
class RBT {
constructor() {
this.root = null;
this.size = 0;
}
isRed(node) {
if (!node) return BLACK;
return node.color;
}
// 左旋 右紅左黑
leftRotate(node) {
let tmp = node.right;
node.right = tmp.left;
tmp.left = node;
tmp.color = node.color;
node.color = RED;
return tmp;
}
// 右旋轉 左紅左子紅
rightRoate(node) {
let tmp = node.left;
node.left = tmp.right;
tmp.right = node;
tmp.color = node.color;
node.color = RED;
return tmp;
}
// 顏色翻轉
flipColors(node) {
node.color = RED;
node.left.color = BLACK;
node.right.color = BLACK;
}
add(key, value) {
this.root = this.addRoot(this.root, key, value);
this.root.color = BLACK; // 根節點始終是黑色
}
addRoot(node, key, value) {
if (!node) {
this.size++;
return new Node(key, value);
}
if (key < node.key) {
node.left = this.addRoot(node.left, key, value);
} else if (key > node.key) {
node.right = this.addRoot(node.right, key, value);
} else {
node.value = value;
}
if (this.isRed(node.right) && !this.isRed((node.left))) {
node = this.leftRotate(node);
}
if (this.isRed(node.left) && this.isRed((node.left.left))) {
node = this.rightRoate(node);
}
if (this.isRed(node.left) && this.isRed(node.right)) {
this.flipColors(node);
}
return node;
}
isEmpty() {
return this.size == 0 ? true : false;
}
getSize() {
return this.size;
}
contains(key) {
let ans = ``;
!(function getNode(node, key) {
if (!node || key == node.key) {
ans = node;
return node;
} else if (key > node.key) {
return getNode(node.right, key);
} else {
return getNode(node.right, key);
}
})(this.root, key);
return !!ans;
}
// bst前序遍歷(遞迴版本)
preOrder(node = this.root) {
if (node == null) return;
console.log(node.key);
this.preOrder(node.left);
this.preOrder(node.right);
}
preOrderNR() {
if (this.root == null) return;
let stack = [];
stack.push(this.root);
while (stack.length > 0) {
let curNode = stack.pop();
console.log(curNode.key);
if (curNode.right != null) stack.push(curNode.right);
if (curNode.left != null) curNode.push(curNode.left);
}
}
// bst中序遍歷
inOrder(node = this.root) {
if (node == null) return;
this.inOrder(node.left);
console.log(node.key);
this.inOrder(node.right);
}
// bst後續遍歷
postOrder(node = this.root) {
if (node == null) return;
this.postOrder(node.left);
this.postOrder(node.right);
console.log(node.key);
}
// bsf + 佇列的方式實現層次遍歷
generateDepthString1() {
let queue = [];
queue.unshift(this.root);
while (queue.length > 0) {
let tmpqueue = []; let ans = [];
queue.forEach(item => {
ans.push(item.key);
item.left ? tmpqueue.push(item.left) : ``;
item.right ? tmpqueue.push(item.right) : ``;
});
console.log(...ans);
queue = tmpqueue;
}
}
minmun(node = this.root) {
if (node.left == null) return node;
return this.minmun(node.left);
}
maximum(node = this.root) {
if (node.right == null) return node;
return this.maximum(node.right);
}
}
let btins = new RBT();
let ary = [5, 3, 6, 8, 4, 2];
ary.forEach(value => btins.add(value));
btins.generateDepthString1();
// ///////////////
// 5 //
// / //
// 3 8 //
// / / //
// 2 4 6 //
// ///////////////
console.log(btins.minmun()); // 2
console.log(btins.maximum()); // 8
圖解