javascript - javascript二叉搜索树实现

Question

任何人都知道 Javascript 中简单 BTree 实现的任何好的例子吗？我有一堆“东西”随机到达，并希望有效地插入每一个。

最终，每个新的都将根据它在树中的最终位置插入到 DOM 中。

我可以从头开始编写代码，但不想重新发明任何轮子。

谢谢

Answer 1

如果重要的话，我发现将这种数据存储为文字树的效率低于将其存储为已排序的数组并在数组上进行二进制搜索以拼接/插入元素。JavaScript 对象的创建显然不是免费的。

还有 ol' encode-a-tree-in-an-array 技巧：

[5, 3, 7, 1, null, 6, 9, null, null, null, null, null, null] 

是相同的

      5
     / \
    3   7
   /   / \
  1   6   9

即 children(N[i]) = N[2i+1], N[2i+2] 。我不知道这是否真的能让你在 JavaScript 中获胜。

如果你尝试一些二叉树的替代方案，你能在这里发布你的发现吗？:)

Answer 2

https://gist.github.com/alexhawkins/f993569424789f3be5db

function BinarySearchTree() {
    this.root = null;
}

BinarySearchTree.prototype.makeNode = function(value) {
    var node = {};
    node.value = value;
    node.left = null;
    node.right = null;
    return node;
};

BinarySearchTree.prototype.add = function(value) {
    var currentNode = this.makeNode(value);
    if (!this.root) {
        this.root = currentNode;
    } else {
        this.insert(currentNode);
    }
    return this;
};

BinarySearchTree.prototype.insert = function(currentNode) {
    var value = currentNode.value;
    var traverse = function(node) {
        //if value is equal to the value of the node, ignore
        //and exit function since we don't want duplicates
        if (value === node.value) {
            return;
        } else if (value > node.value) {
            if (!node.right) {
                node.right = currentNode;
                return;
            } else
                traverse(node.right);
        } else if (value < node.value) {
            if (!node.left) {
                node.left = currentNode;
                return;
            } else
                traverse(node.left);
        }
    };
    traverse(this.root);
};


BinarySearchTree.prototype.contains = function(value) {
    var node = this.root;
    var traverse = function(node) {
        if (!node) return false;
        if (value === node.value) {
            return true;
        } else if (value > node.value) {
            return traverse(node.right);
        } else if (value < node.value) {
            return traverse(node.left);
        }
    };
    return traverse(node);
};


/* BREADTH FIRST TREE TRAVERSAL */

/* Breadth First Search finds all the siblings at each level
   in order from left to right or from right to left. */

BinarySearchTree.prototype.breadthFirstLTR = function() {
    var node = this.root;
    var queue = [node];
    var result = [];
    while (node = queue.shift()) {
        result.push(node.value);
        node.left && queue.push(node.left);
        node.right && queue.push(node.right);
    }
    return result;
};


BinarySearchTree.prototype.breadthFirstRTL = function() {
    var node = this.root;
    var queue = [node];
    var result = [];
    while (node = queue.shift()) {
        result.push(node.value);
        node.right && queue.push(node.right);
        node.left && queue.push(node.left);
    }
    return result;
};

/*DEPTH FIRST TRAVERSALS*/

/*  preOrder is a type of depth-first traversal that tries 
    togo deeper in the tree before exploring siblings. It 
    returns the shallowest descendants first.

    1) Display the data part of root element (or current element)
    2) Traverse the left subtree by recursively calling the pre-order function.
    3) Traverse the right subtree by recursively calling the pre-order function. */

BinarySearchTree.prototype.preOrder = function() {
    var result = [];
    var node = this.root;
    var traverse = function(node) {
        result.push(node.value);
        node.left && traverse(node.left);
        node.right && traverse(node.right);
    };
    traverse(node);
    return result;
};

/*  inOrder traversal is a type of depth-first traversal
    that also tries to go deeper in the tree before exploring siblings.
    however, it returns the deepest descendents first

    1) Traverse the left subtree by recursively calling the pre-order function.
    2) Display the data part of root element (or current element)
    3) Traverse the right subtree by recursively calling the pre-order function. */

BinarySearchTree.prototype.inOrder = function() {
    var result = [];
    var node = this.root;
    var traverse = function(node) {
        node.left && traverse(node.left);
        result.push(node.value);
        node.right && traverse(node.right);
    };
    traverse(node);
    return result;
};

/*  postOrder traversal is a type of depth-first traversal
    that also tries to go deeper in the tree before exploring siblings.
    however, it returns the deepest descendents first

    1) Traverse the left subtree by recursively calling the pre-order function.
    2) Display the data part of root element (or current element)
    3) Traverse the right subtree by recursively calling the pre-order function. */


BinarySearchTree.prototype.postOrder = function() {
    var result = [];
    var node = this.root;
    var traverse = function(node) {
        node.left && traverse(node.left);
        node.right && traverse(node.right);
        result.push(node.value);
    };
    traverse(node);
    return result;
};

//find the left most node to find the min value of a binary tree;
BinarySearchTree.prototype.findMin = function() {
    var node = this.root;
    var traverse = function(node) {
        return !node.left ? node.value : traverse(node.left);
    };
    return traverse(node);
};

//find the right most node to find the max value of a binary tree;
BinarySearchTree.prototype.findMax = function() {
    var node = this.root;
    var traverse = function(node) {
        return !node.right ? node.value : traverse(node.right);
    };
    return traverse(node);
};


BinarySearchTree.prototype.getDepth = function() {
    var node = this.root;
    var maxDepth = 0;
    var traverse = function(node, depth) {
        if (!node) return null;
        if (node) {
            maxDepth = depth > maxDepth ? depth : maxDepth;
            traverse(node.left, depth + 1);
            traverse(node.right, depth + 1);
        }
    };
    traverse(node, 0);
    return maxDepth;
};


//Can you write me a function that returns all the averages of the nodes 
//at each level (or depth)?? with breadth-first traversal


BinarySearchTree.prototype.nodeAverages = function() {
    var node = this.root;
    var result = {};
    var depthAverages = [];

    var traverse = function(node, depth) {
        if (!node) return null;
        if (node) {
            if (!result[depth])
                result[depth] = [node.value];
            else
                result[depth].push(node.value);
        }
        //check to see if node is a leaf, depth stays the same if it is
        //otherwise increment depth for possible right and left nodes
        if (node.right || node.left) {
            traverse(node.left, depth + 1);
            traverse(node.right, depth + 1);
        }
    };
    traverse(node, 0);

    //get averages and breadthFirst
    for (var key in result) {
        var len = result[key].length;
        var depthAvg = 0;
        for (var i = 0; i < len; i++) {
            depthAvg += result[key][i];
        }
        depthAverages.push(Number((depthAvg / len).toFixed(2)));
    }
    return depthAverages;
};

//Convert a binary search tree to a linked-list in place. 
//In-order depth-first traversal.
function LinkedList() {
    this.head = null;
}

BinarySearchTree.prototype.convertToLinkedList = function() {

    var result = [];
    var node = this.root;
    if (!node) return null;

    var traverse = function(node) {
        node.left && traverse(node.left);
        result.push(node.value);
        node.right && traverse(node.right);
    };

    traverse(node);

    var makeNode = function(value) {
        var node = {};
        node.value = value;
        node.next = null;
        return node;
    };

    var list = new LinkedList();
    list.head = makeNode(result[0]);
    var current = list.head;

    for (var i = 1; i < result.length; i++) {
        var currentNode = makeNode(result[i]);
        current.next = currentNode;
        current = current.next;
    }
    return list;
};

//TESTS

var bst = new BinarySearchTree();
bst.add(40).add(25).add(78).add(10).add(32);
console.log('BS1', bst);

var bst2 = new BinarySearchTree();
bst2.add(10).add(20).add(30).add(5).add(8).add(3).add(9);
console.log('BST2', bst2);
console.log('BREADTHFIRST LTR', bst2.breadthFirstLTR());
console.log('BREADTHFIRST RTL', bst2.breadthFirstRTL());
console.log('PREORDER', bst2.preOrder());
console.log('INORDER', bst2.inOrder());
console.log('POSTORDER', bst2.postOrder());

/* 
BREADTHFIRST LTR [ 10, 5, 20, 3, 8, 30, 9 ]
BREADTHFIRST RTL [ 10, 20, 5, 30, 8, 3, 9 ]
PREORDER [ 10, 5, 3, 8, 9, 20, 30 ]
INORDER [ 3, 5, 8, 9, 10, 20, 30 ]
POSTORDER [ 3, 9, 8, 5, 30, 20, 10 ]
*/

var bst3 = new BinarySearchTree();
bst3.add('j').add('f').add('k').add('z').add('a').add('h').add('d');
console.log(bst3);
console.log('BREADTHFIRST LTR', bst3.breadthFirstLTR());
console.log('BREADTHFIRST RTL', bst3.breadthFirstRTL());
console.log('PREORDER', bst3.preOrder());
console.log('INORDER', bst3.inOrder());
console.log('POSTORDER', bst3.postOrder());

/*
BREADTHFIRST LTR [ 'j', 'f', 'k', 'a', 'h', 'z', 'd' ]
BREADTHFIRST RTL [ 'j', 'k', 'f', 'z', 'h', 'a', 'd' ]
PREORDER [ 'j', 'f', 'a', 'd', 'h', 'k', 'z' ]
INORDER [ 'a', 'd', 'f', 'h', 'j', 'k', 'z' ]
POSTORDER [ 'd', 'a', 'h', 'f', 'z', 'k', 'j' ]
 */


console.log(bst2.findMin()); // 3
console.log(bst2.findMax()); // 30
console.log(bst2.contains(15));
//bst2.add(55);
//bst2.add(65);
//bst3.add(75);
console.log(bst2);
console.log(bst2.getDepth()); // 3
console.log(bst2.add(7).add(50).add(80).add(98));
console.log(bst2.getDepth()); // 5
console.log(bst2.nodeAverages()); //[ 10, 12.5, 13.67, 22, 80, 98 ]

console.log(bst2.convertToLinkedList());
//[ 3, 5, 7, 8, 9, 10, 20, 30, 50, 80, 98 ]
//{ head: { value: 3, next: { value: 5, next: [Object] } } }

Answer 3

这有帮助吗？- JavaScript 中的计算机科学：二叉搜索树，第 1 部分

Answer 4

你可以试试buckets，一个 JavaScript 库，它有你需要的一切。

Answer 5

二叉搜索树通常被称为 BST 是一种特殊类型的树，在自然界中是排序的。在二叉搜索树中，每个节点都大于其左子节点且小于其右子节点。此功能使从二叉搜索树中搜索、插入和删除节点变得容易。

算法

//检查根节点是否为空

// 如果是，则将新节点分配给根

// 如果不是，则迭代。Iterate 方法将从当前处理节点的左右子节点检查要添加的节点值。

// 如果要添加的节点值小于该节点而不是移动左孩子

// 如果左子节点为空，则将新节点分配给该左子节点

// 否则调用迭代方法

// 如果要添加的节点值大于节点值而不是移动到右孩子

// 如果右子节点为空，则将新节点分配给该右子节点

// 否则调用迭代方法

如果这对您有帮助，您可以在此处查找完整的文章Binary Search Tree Insert node Implementation in Javascript