• 二叉查找树的实现(中序、前序、后序遍历的实现)


    二叉树和二叉查找树的区别:

    二叉树:每个节点的子节点不允许超过两个。 

    二叉查找树:每个节点的子节点不允许超过两个,同时相对较小的值保存在左节点中, 较大的值保存在右节点中。

    关于中序、前序、后序遍历的理解:

      以从整体角度到分支的角度进行考虑,假设一个小的二叉查找树有三个节点,父节点、左子节点、右子节点,中序遍历是左子节点、父节点、右子节点;前序是父节点、左子节点、右子节点;后序是左子节点、右子节点、父节点。

        function Node(data, left, right) {
            this.data = data;
            this.left = left;
            this.right = right;
            this.show = show;
        }
        function show() {
            return this.data;
        }
        function BST() {
            this.root = null;
            this.insert = insert;
        }
        function insert(data) {
            var n = new Node(data, null, null);
            if (this.root == null) {
                this.root = n;
            } else {
                var current = this.root;
                var parent;
                while (true) {
                    parent = current;
                    if (data < current.data) {
                        current = current.left;
                        if (current == null) {
                            parent.left = n;
                            break;
                        }
                    } else {
                        current = current.right;
                        if (current == null) {
                            parent.right = n;
                            break;
                        }
                    }
                }
            }
        }
        function inOrder(node) {
            if (node != null) {
                inOrder(node.left);
                document.write(node.show() + " ");
                inOrder(node.right);
            }
        }
        var nums = new BST();
        nums.insert(23);
        nums.insert(45);
        nums.insert(16);
        nums.insert(37);
        nums.insert(3);
        nums.insert(99);
        nums.insert(22);
        document.write("Inorder traversal: " + "<br />");
        inOrder(nums.root);
        document.write("<br />");
    
        function preOrder(node) {
            if (node != null) {
                document.write(node.show() + " ");
                preOrder(node.left);
                preOrder(node.right);
            }
        }
        document.write("Preorder traversal: " + "<br />");
        preOrder(nums.root);
        document.write("<br />");
    
        function postOrder(node) {
            if (node != null) {
                postOrder(node.left);
                postOrder(node.right);
                document.write(node.show() + " ");
            }
        }
        document.write("Postorder traversal: " + "<br />");
        postOrder(nums.root);
        /*上述程序运行结果为:
        Inorder traversal: 
        3 16 22 23 37 45 99 
        Preorder traversal: 
        23 16 3 22 45 37 99 
        Postorder traversal: 
        3 22 16 37 99 45 23 */
  • 相关阅读:
    最短路径问题/Spfa
    cddiv/数组维护
    cfdiv2/c/找规律
    Codeforces Round #343 (Div. 2)【A,B水题】
    POJ 2135_Farm Tour
    POJ 3469_Dual Core CPU
    POJ 3469_Dual Core CPU
    POJ 3281_Dining
    POJ 3281_Dining
    POJ 3041_Asteroids
  • 原文地址:https://www.cnblogs.com/feile/p/5391492.html
Copyright © 2020-2023  润新知