给定二叉搜索树，任务是找到具有最小值的节点。
例子：
Input:
Output: 4
方法：只需递归地从根到节点遍历该节点，直到left为NULL。左侧为NULL的节点是具有最小值的节点。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct node {
int data;
struct node* left;
struct node* right;
};
/* Helper function that allocates a new node
with the given data and NULL left and right
pointers. */
struct node* newNode(int data)
{
struct node* node = (struct node*)
malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
return (node);
}
/* Give a binary search tree and a number,
inserts a new node with the given number in
the correct place in the tree. Returns the new
root pointer which the caller should then use
(the standard trick to avoid using reference
parameters). */
struct node* insert(struct node* node, int data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == NULL)
return (newNode(data));
else {
/* 2. Otherwise, recur down the tree */
if (data <= node->data)
node->left = insert(node->left, data);
else
node->right = insert(node->right, data);
/* return the (unchanged) node pointer */
return node;
}
}
// Function to return the minimum node
// in the given binary search tree
int minValue(struct node* node)
{
if (node->left == NULL)
return node->data;
return minValue(node->left);
}
// Driver code
int main()
{
// Create the BST
struct node* root = NULL;
root = insert(root, 4);
insert(root, 2);
insert(root, 1);
insert(root, 3);
insert(root, 6);
insert(root, 5);
cout << minValue(root);
return 0;
} Java
// Java Implementation of the above approach
class GFG
{
/* A binary tree node has data, pointer to left child
and a pointer to right child */
static class Node
{
int data;
Node left;
Node right;
};
/* Helper function that allocates a new node
with the given data and null left and right
pointers. */
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = null;
node.right = null;
return (node);
}
/* Give a binary search tree and a number,
inserts a new node with the given number in
the correct place in the tree. Returns the new
root pointer which the caller should then use
(the standard trick to avoid using reference
parameters). */
static Node insert(Node node, int data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == null)
return (newNode(data));
else
{
/* 2. Otherwise, recur down the tree */
if (data <= node.data)
node.left = insert(node.left, data);
else
node.right = insert(node.right, data);
/* return the (unchanged) node pointer */
return node;
}
}
// Function to return the minimum node
// in the given binary search tree
static int minValue(Node node)
{
if (node.left == null)
return node.data;
return minValue(node.left);
}
// Driver code
public static void main(String args[])
{
// Create the BST
Node root = null;
root = insert(root, 4);
insert(root, 2);
insert(root, 1);
insert(root, 3);
insert(root, 6);
insert(root, 5);
System.out.println(minValue(root));
}
}
// This code has been contributed by 29AjayKumarPython3
# Python3 Implementation of
# the above approach
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Helper function that allocates
# a new node with the given data
# and null left and right pointers.
def insert(node, data):
if node is None :
return Node(data)
else:
if data <= node.data:
node.left = insert(node.left, data)
else:
node.right = insert(node.right, data)
return node
# Function to return the minimum node
# in the given binary search tree
def minValue(node):
if node.left == None:
return node.data
return minValue(node.left)
# Driver code
if __name__ == "__main__" :
# Create the BST
root = None
root = insert(root, 4)
insert(root, 2)
insert(root, 1)
insert(root, 3)
insert(root, 6)
insert(root, 5)
print(minValue(root))
# This code is contributed by vinayakC#
// C# Implementation of the above approach
using System;
class GFG
{
/* A binary tree node has data, pointer to left child
and a pointer to right child */
public class Node
{
public int data;
public Node left;
public Node right;
};
/* Helper function that allocates a new node
with the given data and null left and right
pointers. */
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = null;
node.right = null;
return (node);
}
/* Give a binary search tree and a number,
inserts a new node with the given number in
the correct place in the tree. Returns the new
root pointer which the caller should then use
(the standard trick to avoid using reference
parameters). */
static Node insert(Node node, int data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == null)
return (newNode(data));
else
{
/* 2. Otherwise, recur down the tree */
if (data <= node.data)
node.left = insert(node.left, data);
else
node.right = insert(node.right, data);
/* return the (unchanged) node pointer */
return node;
}
}
// Function to return the minimum node
// in the given binary search tree
static int minValue(Node node)
{
if (node.left == null)
return node.data;
return minValue(node.left);
}
// Driver code
public static void Main(String []args)
{
// Create the BST
Node root = null;
root = insert(root, 4);
insert(root, 2);
insert(root, 1);
insert(root, 3);
insert(root, 6);
insert(root, 5);
Console.WriteLine(minValue(root));
}
}
// This code contributed by Rajput-Ji输出：
1
时间复杂度： O(n)，最坏的情况发生在左偏斜的树上。