二叉查找树的常用操作

1、结点的构造

struct BSTnode {
	int key;
	BSTnode* left;
	BSTnode* right;
	BSTnode(int K) {
		key = K;
		left = right = NULL;
	}
};

2、查找算法

2.1 递归版本

BSTnode* Search(BSTnode* root, int K) {
	if (root == NULL || root->key == K) return root;
	if (K < root->key) return Search(root->left, K);
	else return Search(root->right, K);
}

例题：LeetCode 700

2.2 迭代版本

BSTnode* Search2(BSTnode* root, int K) {
	BSTnode* p = root;
	while (p != NULL) {
		if (K < p->key) p = p->left;
		else if (K > p->key) p = p->right;
		else return p;
	}
	return NULL;
}

3、插入算法

3.1 引用传值

void insert(BSTnode* &root, int K) {
	if (root == NULL) root = new BSTnode(K);
	else if (K < root->key) insert(root->left, K);
	else if (K > root->key) insert(root->right, K);
}

3.2 值传递

BSTnode* Insert(BSTnode* root, int K) {
	if (root == NULL) root = new BSTnode(K);
	else if (K < root->key) root->left = Insert(root->left, K);
	else if (K > root->key) root->right = Insert(root->right, K);
	return root;
}

例题：LeetCode 701

4、删除算法

4.1 引用传值

void remove(BSTnode* &root, int K) {
	if (root == NULL) return;
	if (K < root->key) remove(root->left, K);//在左子树删K
	else if (K > root->key) remove(root->right, K);//在右子树删K
	else if (root->left != NULL && root->right != NULL) {
		BSTnode* s = root->right;
		while(s->left != NULL){
			s = s->left;
		}
		root->key = s->key;//s为t右子树中根序列第一个节点
		remove(root->right, s->key);
	}
	else {
		BSTnode* oldroot = root;
		root = (root->left != NULL) ? root->left : root->right;
		delete oldroot;
	}
}

4.2 值传递

BSTnode* deleteNode(BSTnode* root, int K) {
	if (root == NULL) return root;
	if (K < root->key) {
		root->left = deleteNode(root->left, K);
	}
	else if (K > root->key) {
		root->right = deleteNode(root->right, K);
	}
	else if (root->left != NULL && root->right != NULL) {
		BSTnode* s = root->right;
		while(s->left != NULL){
			s = s->left;
		}
		root->key = s->key;
		root->right = deleteNode(root->right, s->key);
	}
	else {
		BSTnode* oldroot = root;
		root = (root->left != NULL) ? root->left : root->right;
		delete oldroot;
	}
	return root;
}

例题：LeetCode 450

5、统计由n个结点组成且结点关键词为1到n的二叉查找树有多少种

int numTrees(int n) {
	long long C = 1;
	for (int i = 2; i <= n; i++) {
		C = C * 2 * (2 * i - 1) / (i + 1);
	}
	return C;
}

例题：LeetCode 96