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	* ci(kotlin): Add workflow file. * Update kotlin.yml * style(kotlin): value -> _val --------- Co-authored-by: Yudong Jin <krahets@163.com>
		
			
				
	
	
		
			157 lines
		
	
	
		
			4.4 KiB
		
	
	
	
		
			Kotlin
		
	
	
	
	
	
			
		
		
	
	
			157 lines
		
	
	
		
			4.4 KiB
		
	
	
	
		
			Kotlin
		
	
	
	
	
	
/**
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 * File: binary_search_tree.kt
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 * Created Time: 2024-01-25
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 * Author: curtishd (1023632660@qq.com)
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 */
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package chapter_tree
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import utils.TreeNode
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import utils.printTree
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/* 二叉搜索树 */
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class BinarySearchTree {
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    // 初始化空树
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    private var root: TreeNode? = null
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    /* 获取二叉树根节点 */
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    fun getRoot(): TreeNode? {
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        return root
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    }
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    /* 查找节点 */
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    fun search(num: Int): TreeNode? {
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        var cur = root
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        // 循环查找,越过叶节点后跳出
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        while (cur != null) {
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            // 目标节点在 cur 的右子树中
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            cur = if (cur._val < num)
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                cur.right
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            // 目标节点在 cur 的左子树中
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            else if (cur._val > num)
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                cur.left
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            // 找到目标节点,跳出循环
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            else
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                break
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        }
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        // 返回目标节点
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        return cur
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    }
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    /* 插入节点 */
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    fun insert(num: Int) {
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        // 若树为空,则初始化根节点
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        if (root == null) {
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            root = TreeNode(num)
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            return
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        }
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        var cur = root
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        var pre: TreeNode? = null
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        // 循环查找,越过叶节点后跳出
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        while (cur != null) {
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            // 找到重复节点,直接返回
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            if (cur._val == num)
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                return
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            pre = cur
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            // 插入位置在 cur 的右子树中
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            cur = if (cur._val < num)
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                cur.right
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            // 插入位置在 cur 的左子树中
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            else
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                cur.left
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        }
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        // 插入节点
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        val node = TreeNode(num)
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        if (pre?._val!! < num)
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            pre.right = node
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        else
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            pre.left = node
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    }
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    /* 删除节点 */
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    fun remove(num: Int) {
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        // 若树为空,直接提前返回
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        if (root == null)
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            return
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        var cur = root
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        var pre: TreeNode? = null
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        // 循环查找,越过叶节点后跳出
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        while (cur != null) {
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            // 找到待删除节点,跳出循环
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            if (cur._val == num)
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                break
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            pre = cur
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            // 待删除节点在 cur 的右子树中
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            cur = if (cur._val < num)
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                cur.right
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            // 待删除节点在 cur 的左子树中
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            else
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                cur.left
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        }
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        // 若无待删除节点,则直接返回
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        if (cur == null)
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            return
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        // 子节点数量 = 0 or 1
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        if (cur.left == null || cur.right == null) {
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            // 当子节点数量 = 0 / 1 时, child = null / 该子节点
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            val child = if (cur.left != null)
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                cur.left
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            else
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                cur.right
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            // 删除节点 cur
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            if (cur != root) {
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                if (pre!!.left == cur)
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                    pre.left = child
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                else
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                    pre.right = child
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            } else {
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                // 若删除节点为根节点,则重新指定根节点
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                root = child
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            }
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            // 子节点数量 = 2
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        } else {
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            // 获取中序遍历中 cur 的下一个节点
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            var tmp = cur.right
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            while (tmp!!.left != null) {
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                tmp = tmp.left
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            }
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            // 递归删除节点 tmp
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            remove(tmp._val)
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            // 用 tmp 覆盖 cur
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            cur._val = tmp._val
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        }
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    }
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}
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/* Driver Code */
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fun main() {
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    /* 初始化二叉搜索树 */
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    val bst = BinarySearchTree()
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    // 请注意,不同的插入顺序会生成不同的二叉树,该序列可以生成一个完美二叉树
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    val nums = intArrayOf(8, 4, 12, 2, 6, 10, 14, 1, 3, 5, 7, 9, 11, 13, 15)
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    for (num in nums) {
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        bst.insert(num)
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    }
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    println("\n初始化的二叉树为\n")
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    printTree(bst.getRoot())
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    /* 查找节点 */
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    val node = bst.search(7)
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    println("查找到的节点对象为 $node,节点值 = ${node?._val}")
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    /* 插入节点 */
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    bst.insert(16)
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    println("\n插入节点 16 后,二叉树为\n")
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    printTree(bst.getRoot())
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    /* 删除节点 */
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    bst.remove(1)
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    println("\n删除节点 1 后,二叉树为\n")
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    printTree(bst.getRoot())
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    bst.remove(2)
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    println("\n删除节点 2 后,二叉树为\n")
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    printTree(bst.getRoot())
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    bst.remove(4)
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    println("\n删除节点 4 后,二叉树为\n")
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    printTree(bst.getRoot())
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} |