给定一个由N 个节点和两个节点A和B组成的二叉搜索树 (BST)，任务是找到给定 BST 中所有节点的中值，这些节点的值位于[A, B]范围内。
例子：
Input: A = 3, B = 11
_0.jpg)
Output: 6
Explanation:
The nodes that lie over the range [3, 11] are {3, 4, 6, 8, 11}. The median of the given nodes is 6.
Input: A = 6, B = 15
_1.jpg)
Output: 9.5
方法：可以通过对给定树执行任意树遍历并存储位于[A, B]范围内的所有节点来解决给定问题，并找到所有存储元素的中值。请按照以下步骤解决问题：
- 初始化一个向量，比如V ，它存储位于[A, B]范围内的树的所有值。
- 执行给定树的中序遍历，如果任何节点的值位于[A, B]范围内，则将该值插入向量V 中。
- 完成上述步骤后，将存储在向量V中的所有元素的中值作为结果打印出来。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Tree Node structure
struct Node {
struct Node *left, *right;
int key;
};
// Function to create a new BST node
Node* newNode(int key)
{
Node* temp = new Node;
temp->key = key;
temp->left = temp->right = NULL;
return temp;
}
// Function to insert a new node with
// given key in BST
Node* insertNode(Node* node, int key)
{
// If the tree is empty,
// return a new node
if (node == NULL)
return newNode(key);
// Otherwise, recur down the tree
if (key < node->key)
node->left = insertNode(
node->left, key);
else if (key > node->key)
node->right = insertNode(
node->right, key);
// Return the node pointer
return node;
}
// Function to find all the nodes that
// lies over the range [node1, node2]
void getIntermediateNodes(
Node* root, vector& interNodes,
int node1, int node2)
{
// If the tree is empty, return
if (root == NULL)
return;
// Traverse for the left subtree
getIntermediateNodes(root->left,
interNodes,
node1, node2);
// If a second node is found,
// then update the flag as false
if (root->key <= node2
and root->key >= node1) {
interNodes.push_back(root->key);
}
// Traverse the right subtree
getIntermediateNodes(root->right,
interNodes,
node1, node2);
}
// Function to find the median of all
// the values in the given BST that
// lies over the range [node1, node2]
float findMedian(Node* root, int node1,
int node2)
{
// Stores all the nodes in
// the range [node1, node2]
vector interNodes;
getIntermediateNodes(root, interNodes,
node1, node2);
// Store the size of the array
int nSize = interNodes.size();
// Print the median of array
// based on the size of array
return (nSize % 2 == 1)
? (float)interNodes[nSize / 2]
: (float)(interNodes[(nSize - 1) / 2]
+ interNodes[nSize / 2])
/ 2;
}
// Driver Code
int main()
{
// Given BST
struct Node* root = NULL;
root = insertNode(root, 8);
insertNode(root, 3);
insertNode(root, 1);
insertNode(root, 6);
insertNode(root, 4);
insertNode(root, 11);
insertNode(root, 15);
cout << findMedian(root, 3, 11);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Tree Node structure
static class Node
{
Node left, right;
int key;
};
static Vector interNodes = new Vector();
// Function to create a new BST node
static Node newNode(int key)
{
Node temp = new Node();
temp.key = key;
temp.left = temp.right = null;
return temp;
}
// Function to insert a new node with
// given key in BST
static Node insertNode(Node node, int key)
{
// If the tree is empty,
// return a new node
if (node == null)
return newNode(key);
// Otherwise, recur down the tree
if (key < node.key)
node.left = insertNode(
node.left, key);
else if (key > node.key)
node.right = insertNode(
node.right, key);
// Return the node pointer
return node;
}
// Function to find all the nodes that
// lies over the range [node1, node2]
static void getIntermediateNodes(Node root,
int node1,
int node2)
{
// If the tree is empty, return
if (root == null)
return;
// Traverse for the left subtree
getIntermediateNodes(root.left,
node1, node2);
// If a second node is found,
// then update the flag as false
if (root.key <= node2 &&
root.key >= node1)
{
interNodes.add(root.key);
}
// Traverse the right subtree
getIntermediateNodes(root.right,
node1, node2);
}
// Function to find the median of all
// the values in the given BST that
// lies over the range [node1, node2]
static float findMedian(Node root, int node1,
int node2)
{
// Stores all the nodes in
// the range [node1, node2]
getIntermediateNodes(root,
node1, node2);
// Store the size of the array
int nSize = interNodes.size();
// Print the median of array
// based on the size of array
return (nSize % 2 == 1) ?
(float)interNodes.get(nSize / 2) :
(float)(interNodes.get((nSize - 1) / 2) +
interNodes.get(nSize / 2)) / 2;
}
// Driver Code
public static void main(String[] args)
{
// Given BST
Node root = null;
root = insertNode(root, 8);
insertNode(root, 3);
insertNode(root, 1);
insertNode(root, 6);
insertNode(root, 4);
insertNode(root, 11);
insertNode(root, 15);
System.out.print(findMedian(root, 3, 11));
}
}
// This code is contributed by shikhasingrajput C#
// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG{
// Tree Node structure
class Node
{
public Node left, right;
public int key;
};
static List interNodes = new List();
// Function to create a new BST node
static Node newNode(int key)
{
Node temp = new Node();
temp.key = key;
temp.left = temp.right = null;
return temp;
}
// Function to insert a new node with
// given key in BST
static Node insertNode(Node node, int key)
{
// If the tree is empty,
// return a new node
if (node == null)
return newNode(key);
// Otherwise, recur down the tree
if (key < node.key)
node.left = insertNode(
node.left, key);
else if (key > node.key)
node.right = insertNode(
node.right, key);
// Return the node pointer
return node;
}
// Function to find all the nodes that
// lies over the range [node1, node2]
static void getIntermediateNodes(Node root,
int node1,
int node2)
{
// If the tree is empty, return
if (root == null)
return;
// Traverse for the left subtree
getIntermediateNodes(root.left,
node1, node2);
// If a second node is found,
// then update the flag as false
if (root.key <= node2 &&
root.key >= node1)
{
interNodes.Add(root.key);
}
// Traverse the right subtree
getIntermediateNodes(root.right,
node1, node2);
}
// Function to find the median of all
// the values in the given BST that
// lies over the range [node1, node2]
static float findMedian(Node root, int node1,
int node2)
{
// Stores all the nodes in
// the range [node1, node2]
getIntermediateNodes(root,
node1, node2);
// Store the size of the array
int nSize = interNodes.Count;
// Print the median of array
// based on the size of array
return (nSize % 2 == 1) ?
(float)interNodes[nSize / 2] :
(float)(interNodes[(nSize - 1) / 2] +
interNodes[nSize / 2]) / 2;
}
// Driver Code
public static void Main(String[] args)
{
// Given BST
Node root = null;
root = insertNode(root, 8);
insertNode(root, 3);
insertNode(root, 1);
insertNode(root, 6);
insertNode(root, 4);
insertNode(root, 11);
insertNode(root, 15);
Console.Write(findMedian(root, 3, 11));
}
}
// This code is contributed by shikhasingrajput 输出：
6时间复杂度： O(N)
辅助空间： O(N)
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