Java学习苦旅（二十三）—

Java学习苦旅（二十三）——二叉搜索树

news2024/11/28 6:41:56

本篇博客将详细讲解二叉搜索树。

文章目录

二叉搜索树
- 概念
- 操作
- - 查找
  - 插入
  - 删除
- 性能分析
结尾

二叉搜索树

概念

二叉搜索树又称二叉排序树，它或者是一棵空树，或者是具有以下性质的二叉树:

若它的左子树不为空，则左子树上所有节点的值都小于根节点的值

若它的右子树不为空，则右子树上所有节点的值都大于根节点的值

它的左右子树也分别为二叉搜索树

操作

查找

示例代码

class Node {
    public int val;
    public Node left;
    public Node right;

    public Node(int val) {
        this.val = val;
    }
}

public class BinarySearchTree {
    public Node root = null;

    public Node search(int key) {
        Node cur = root;
        while (cur != null) {
            if (cur.val < key) {
                cur = cur.right;
            } else if (cur.val == key) {
                return cur;
            } else {
                cur = cur.left;
            }
        }
        return null;
    }
}

插入

具体步骤：

如果树为空树，即根 == null，直接插入
如果树不是空树，按照查找逻辑确定插入位置，插入新结点

示例代码

public boolean insert(int val) {
    if (root == null) {
        root = new Node(val);
        return true;
    }
    Node cur = root;
    Node parent = null;
    while (cur != null) {
        if (cur.val < val) {
            parent = cur;
            cur = cur.right;
        } else if (cur.val == val) {
            return false;//不能有相同的数据
        } else {
            parent = cur;
            cur = cur.left;
        }
    }
    Node node = new Node(val);
    if (parent.val < val) {
        parent.right = node;
    } else {
        parent.left = node;
    }
    return true;
}

删除

设待删除结点为 cur, 待删除结点的双亲结点为parent

cur.left == null

cur 是 root，则 root = cur.right
cur 不是 root，cur 是 parent.left，则 parent.left = cur.right
cur 不是 root，cur 是 parent.right，则 parent.right = cur.right

cur.right == null

cur 是 root，则 root = cur.left
cur 不是 root，cur 是 parent.left，则 parent.left = cur.left
cur 不是 root，cur 是 parent.right，则 parent.right = cur.left

cur.left != null && cur.right != null

需要使用替换法进行删除，即在它的右子树中寻找中序下的第一个结点(关键码最小)，用它的值填补到被删除节点中，再来处理该结点的删除问题

示例代码

public void remove(int key) {
    Node cur = root;
    Node parent = null;
    while (cur != null) {
        if (cur.val == key) {
            removeNode(cur,parent);
            break;
        }else if (cur.val < key) {
            parent = cur;
            cur = cur.right;
        } else {
            parent = cur;
            cur = cur.left;
        }
    }
}

public void removeNode(Node cur, Node parent) {
    if (cur.left == null) {
        if (cur == root) {
            root = cur.right;
        } else if (cur == parent.left) {
            parent.left = cur.right;
        } else {
            parent.right = cur.right;
        }
    } else if (cur.right == null) {
        if (cur == root) {
            root = cur.left;
        } else if (cur == parent.left) {
            parent.left = cur.left;
        } else {
            parent.right = cur.left;
        }
    } else {
        Node targetParent = cur;
        Node target = cur.right;
        while (target.left != null) {
            targetParent = target;
            target = target.left;
        }
        cur.val = target.val;
        if (target == targetParent.left) {
            targetParent.left = target.right;
        } else {
            targetParent.right = target.right;
        }
    }
}