给定一棵二叉搜索树和二叉搜索树的一个节点，任务是迭代地从二叉搜索树中删除该节点。
以下是对 BST 执行删除操作时出现的三种情况：
1. 情况一：待删除节点为叶节点。直接从树中删除节点。
10 10
/ \ delete(5) / \
7 15 ---------> 7 15
/ \ / \ \ / \
5 8 11 18 8 11 18 2. 情况二：要删除的节点是一个有两个孩子的内部节点。复制待删除节点的中序后继内容，删除中序后继。可以通过在节点的右子树中找到最小元素来找到中序后继。
inorderSuccessor(10) = 11。
10 11
/ \ delete(10) / \
7 15 ---------> 7 15
/ \ / \ / \ \
5 8 11 18 5 8 183. 情况三：待删除节点是一个内部节点，有一个孩子。在这种情况下，删除节点并将其子节点向上移动以代替它。
10 10
/ \ delete(15) / \
7 15 ---------> 7 11
/ \ / / \
5 8 11 5 8在案例 2 中删除中序后继的直觉是具有两个孩子的节点的中序后继将始终大于该节点的左子树中的所有元素，因为它是右子中的最小节点节点的树和节点的中序后继节点将始终小于节点右子树中的所有其他节点。
这保留了给定节点的左子树中所有节点的 BST 属性都小于给定节点，并且给定节点的右子树中的所有节点都大于给定节点。
下面是上述方法的实现：
C++
// C++ implementation to delete
// a node in the BST
#include
using namespace std;
// Structure of the node
typedef struct treeNode {
int data;
struct treeNode* left;
struct treeNode* right;
} treeNode;
// Utility function to print
// the inorder traversal of the BST.
void inorder(treeNode* root)
{
if (root != NULL) {
inorder(root->left);
cout << root->data << ' ';
inorder(root->right);
}
}
// Utility function to insert
// nodes into our BST
treeNode* insert(treeNode* root, int key)
{
// Check if tree is empty
if (root == NULL) {
treeNode* temp;
temp = (treeNode*)malloc(sizeof(treeNode));
temp->data = key;
temp->left = NULL;
temp->right = NULL;
return temp;
}
if (key < root->data) {
// if the key to be inserted
// is lesser than the root,
// insert into the left subtree,
// and recursively call
// the insert function with the
// root->left as the new root.
root->left = insert(root->left, key);
}
else {
// if the key to be inserted
// is greater than the root,
// insert into the right subtree,
// and recursively call
// the insert function with the
// root->right as the new root.
root->right = insert(root->right, key);
}
return root;
}
// Iterative Function to delete
// 'key' from the BST.
treeNode* deleteIterative(treeNode* root, int key)
{
treeNode* curr = root;
treeNode* prev = NULL;
// Check if the key is actually
// present in the BST.
// the variable prev points to
// the parent of the key to be deleted.
while (curr != NULL && curr->data != key) {
prev = curr;
if (key < curr->data)
curr = curr->left;
else
curr = curr->right;
}
if (curr == NULL) {
cout << "Key " << key << " not found in the"
<< " provided BST.\n";
return root;
}
// Check if the node to be
// deleted has atmost one child.
if (curr->left == NULL || curr->right == NULL) {
// newCurr will replace
// the node to be deleted.
treeNode* newCurr;
// if the left child does not exist.
if (curr->left == NULL)
newCurr = curr->right;
else
newCurr = curr->left;
// check if the node to
// be deleted is the root.
if (prev == NULL)
return newCurr;
// check if the node to be deleted
// is prev's left or right child
// and then replace this with newCurr
if (curr == prev->left)
prev->left = newCurr;
else
prev->right = newCurr;
// free memory of the
// node to be deleted.
free(curr);
}
// node to be deleted has
// two children.
else {
treeNode* p = NULL;
treeNode* temp;
// Compute the inorder successor
temp = curr->right;
while (temp->left != NULL) {
p = temp;
temp = temp->left;
}
// check if the parent of the inorder
// successor is the curr or not(i.e. curr=
// the node which has the same data as
// the given data by the user to be
// deleted). if it isn't, then make the
// the left child of its parent equal to
// the inorder successor'd right child.
if (p != NULL)
p->left = temp->right;
// if the inorder successor was the
// curr (i.e. curr = the node which has the
// same data as the given data by the
// user to be deleted), then make the
// right child of the node to be
// deleted equal to the right child of
// the inorder successor.
else
curr->right = temp->right;
curr->data = temp->data;
free(temp);
}
return root;
}
// Driver Code
int main()
{
/*
10
/ \
7 15
/ \ / \
5 8 11 18
*/
treeNode* root = NULL;
root = insert(root, 10);
root = insert(root, 7);
root = insert(root, 5);
root = insert(root, 8);
root = insert(root, 15);
root = insert(root, 11);
root = insert(root, 18);
cout << "Inorder traversal "
<< "of original BST:\n";
inorder(root);
cout << '\n';
// delete node with data value 11 (leaf)
root = deleteIterative(root, 11);
cout << "\nDeletion of 11\n";
cout << "Inorder traversal post deletion:\n";
inorder(root);
cout << '\n';
// delete node with data value 15
// (internal node with one child)
root = deleteIterative(root, 15);
cout << "\nDeletion of 15\n";
cout << "Inorder traversal post deletion:\n";
inorder(root);
cout << '\n';
// delete node with data value 10
// (root, two children)
root = deleteIterative(root, 10);
cout << "\nDeletion of 10\n";
cout << "Inorder traversal post deletion:\n";
inorder(root);
cout << '\n';
return 0;
} Python3
# Python implementation to delete
# a node in the Binary Search Tree
# Class for a node of BST.
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Utility function to print
# the inorder traversal of the BST
def inorder(root):
if root != None:
inorder(root.left)
print(root.data, end=" ")
inorder(root.right)
# Utility function to insert
# nodes into our BST
def insert(root, key):
# check if tree is empty
if root == None:
temp = Node(key)
return temp
if key < root.data:
"""
if the key to be inserted is
lesser than the root,
insert into the left subtree,
and recursively call
the insert function with
the root.left as the new root.
"""
root.left = insert(root.left, key)
else:
"""
if the key to be inserted is
greater than the root,
insert into the right subtree,
and recursively call
the insert function with the
root->right as the new root.
"""
root.right = insert(root.right, key)
return root
# Iterative appraoch to
# delete 'key' from the BST.
def deleteIterative(root, key):
curr = root
prev = None
# First check if the key is
# actually present in the BST.
# the variable prev points to the
# parent of the key to be deleted
while(curr != None and curr.data != key):
prev = curr
if curr.data < key:
curr = curr.right
else:
curr = curr.left
if curr == None:
print("Key % d not found in\
the provided BST." % key)
return root
# Check if the node to be
# deleted has atmost one child
if curr.left == None or\
curr.right == None:
# newCurr will replace
# the node to be deleted.
newCurr = None
# if the left child does not exist.
if curr.left == None:
newCurr = curr.right
else:
newCurr = curr.left
# check if the node to
# be deleted is the root.
if prev == None:
return newCurr
# Check if the node to be
# deleted is prev's left or
# right child and then
# replace this with newCurr
if curr == prev.left:
prev.left = newCurr
else:
prev.right = newCurr
curr = None
# node to be deleted
# has two children.
else:
p = None
temp = None
# Compute the inorder
# successor of curr.
temp = curr.right
while(temp.left != None):
p = temp
temp = temp.left
# check if the parent of the
# inorder successor is the root or not.
# if it isn't, then make the left
# child of its parent equal to the
# inorder successor's right child.
if p != None:
p.left = temp.right
else:
# if the inorder successor was
# the root, then make the right child
# of the node to be deleted equal
# to the right child of the inorder
# successor.
curr.right = temp.right
curr.data = temp.data
temp = None
return root
# Function to create the BST
# and call the Delete Function
def main():
"""
10
/ \
7 15
/ \ / \
5 8 11 18
"""
root = None
root = insert(root, 10)
root = insert(root, 7)
root = insert(root, 5)
root = insert(root, 8)
root = insert(root, 15)
root = insert(root, 11)
root = insert(root, 18)
print("Inorder traversal of original BST:")
inorder(root)
print("\n")
# delete node with data value 11 (leaf)
root = deleteIterative(root, 11)
print("Deletion of 11")
print("Inorder traversal post deletion:")
inorder(root)
print("\n")
# delete node with data value 15
# (internal node with one child)
root = deleteIterative(root, 15)
print("Deletion of 15")
print("Inorder traversal post deletion:")
inorder(root)
print("\n")
# delete node with data value 10
# (root, two children)
root = deleteIterative(root, 10)
print("Deletion of 10")
print("Inorder traversal post deletion:")
inorder(root)
print()
# Driver Code
if __name__ == "__main__":
main()输出：
Inorder traversal of original BST:
5 7 8 10 11 15 18
Deletion of 11
Inorder traversal post deletion:
5 7 8 10 15 18
Deletion of 15
Inorder traversal post deletion:
5 7 8 10 18
Deletion of 10
Inorder traversal post deletion:
5 7 8 18如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live