Question

我正在尝试构建一个看起来像二叉树的结构。我需要在哪里设置后向指针，以便每个节点都可以链接到其父节点，这对性能有何影响？

function Node(expression, trueStatement, falseStatement) {
    this.expression = expression;
    this.trueStatement = trueStatement
    this.falseStatement = falseStatement
    this.left = null;
    this.right = null;
    this.back = null;
}

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

BinarySearchTree.prototype.push = function(state, expression, trueStatement, falseStatement) {
    var root = this.root;

    if (trueStatement == "") {
        trueStatement = false;
    }
    if (falseStatement == "") {
        falseStatement = false;
    }

    if (!root) {
        this.root = new Node(expression, trueStatement, falseStatement);
        return;
    }



    var currentNode = root;
    var newNode = new Node(expression, trueStatement, falseStatement);

    while (currentNode) {
        if (state) {
            if (!currentNode.left) {
                currentNode.left = newNode;
                break;
            } else {
                currentNode = currentNode.left;
            }
        } else {
            if (!currentNode.right) {
                currentNode.right = newNode;
                break;
            } else {
                currentNode = currentNode.right;
            }
        }
    }

}
var bst = new BinarySearchTree();
bst.push(true, "IIF()", "", "@user");
bst.push(false, "IIF()1", "@user", "");
bst.push(true, "IIF()2", "@user", "");

目标是每个节点都应与其父节点链接。

Answer 1

将属性parent / back添加到您的Node对象，并在创建Node时进行填充。以及您在哪里

currentNode.left = newNode;

还添加

 currentNode.left = newNode;
 newNode.back = currentNode;

在其他地方实施相同的逻辑

有没有可能在树结构中向父添加回指针？

1 个答案: