求 BST 在 O(n)时间和 O(1)空间的中位数
原文:https://www . geesforgeks . org/find-median-BST-time-O1-space/
给定一个二叉查找树，求它的中位数。 如果节点数为偶数:则中位数=((n/2 个节点+(n+1)/2 个节点)/2 如果节点数为奇数:则中位数=(n+1)/2 个节点。 例如，英国标准时间以下的中位数是 12。
更多示例:
Given BST(with odd no. of nodes) is :
6
/ \
3 8
/ \ / \
1 4 7 9
Inorder of Given BST will be : 1, 3, 4, 6, 7, 8, 9
So, here median will 6.
Given BST(with even no. of nodes) is :
6
/ \
3 8
/ \ /
1 4 7
Inorder of Given BST will be : 1, 3, 4, 6, 7, 8
So, here median will (4+6)/2 = 5.
问于:谷歌
为了找到中间值，我们需要找到 BST 的 inoder，因为它的 inoder 将按排序顺序排列，然后找到中间值，即 这个想法是基于BST 中的第 K 个最小元素使用 O(1) Extra Space 这个任务非常简单，如果我们被允许使用额外的空间，但是使用递归的 inoder 遍历和堆栈都使用 Space，这在这里是不允许的。因此，解决方案是进行莫里斯无序遍历，因为它不需要任何额外的空间。
Implementation:
1- Count the no. of nodes in the given BST
using Morris Inorder Traversal.
2- Then Perform Morris Inorder traversal one
more time by counting nodes and by checking if
count is equal to the median point.
To consider even no. of nodes an extra pointer
pointing to the previous node is used.
C++
/* C++ program to find the median of BST in O(n)
time and O(1) space*/
#include<bits/stdc++.h>
using namespace std;
/* A binary search tree Node has data, pointer
to left child and a pointer to right child */
struct Node
{
int data;
struct Node* left, *right;
};
// A utility function to create a new BST node
struct Node *newNode(int item)
{
struct Node *temp = new Node;
temp->data = item;
temp->left = temp->right = NULL;
return temp;
}
/* A utility function to insert a new node with
given key in BST */
struct Node* insert(struct 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->data)
node->left = insert(node->left, key);
else if (key > node->data)
node->right = insert(node->right, key);
/* return the (unchanged) node pointer */
return node;
}
/* Function to count nodes in a binary search tree
using Morris Inorder traversal*/
int counNodes(struct Node *root)
{
struct Node *current, *pre;
// Initialise count of nodes as 0
int count = 0;
if (root == NULL)
return count;
current = root;
while (current != NULL)
{
if (current->left == NULL)
{
// Count node if its left is NULL
count++;
// Move to its right
current = current->right;
}
else
{
/* Find the inorder predecessor of current */
pre = current->left;
while (pre->right != NULL &&
pre->right != current)
pre = pre->right;
/* Make current as right child of its
inorder predecessor */
if(pre->right == NULL)
{
pre->right = current;
current = current->left;
}
/* Revert the changes made in if part to
restore the original tree i.e., fix
the right child of predecessor */
else
{
pre->right = NULL;
// Increment count if the current
// node is to be visited
count++;
current = current->right;
} /* End of if condition pre->right == NULL */
} /* End of if condition current->left == NULL*/
} /* End of while */
return count;
}
/* Function to find median in O(n) time and O(1) space
using Morris Inorder traversal*/
int findMedian(struct Node *root)
{
if (root == NULL)
return 0;
int count = counNodes(root);
int currCount = 0;
struct Node *current = root, *pre, *prev;
while (current != NULL)
{
if (current->left == NULL)
{
// count current node
currCount++;
// check if current node is the median
// Odd case
if (count % 2 != 0 && currCount == (count+1)/2)
return prev->data;
// Even case
else if (count % 2 == 0 && currCount == (count/2)+1)
return (prev->data + current->data)/2;
// Update prev for even no. of nodes
prev = current;
//Move to the right
current = current->right;
}
else
{
/* Find the inorder predecessor of current */
pre = current->left;
while (pre->right != NULL && pre->right != current)
pre = pre->right;
/* Make current as right child of its inorder predecessor */
if (pre->right == NULL)
{
pre->right = current;
current = current->left;
}
/* Revert the changes made in if part to restore the original
tree i.e., fix the right child of predecessor */
else
{
pre->right = NULL;
prev = pre;
// Count current node
currCount++;
// Check if the current node is the median
if (count % 2 != 0 && currCount == (count+1)/2 )
return current->data;
else if (count%2==0 && currCount == (count/2)+1)
return (prev->data+current->data)/2;
// update prev node for the case of even
// no. of nodes
prev = current;
current = current->right;
} /* End of if condition pre->right == NULL */
} /* End of if condition current->left == NULL*/
} /* End of while */
}
/* Driver program to test above functions*/
int main()
{
/* Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80 */
struct Node *root = NULL;
root = insert(root, 50);
insert(root, 30);
insert(root, 20);
insert(root, 40);
insert(root, 70);
insert(root, 60);
insert(root, 80);
cout << "\nMedian of BST is "
<< findMedian(root);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
/* Java program to find the median of BST in O(n)
time and O(1) space*/
class GfG {
/* A binary search tree Node has data, pointer
to left child and a pointer to right child */
static class Node
{
int data;
Node left, right;
}
// A utility function to create a new BST node
static Node newNode(int item)
{
Node temp = new Node();
temp.data = item;
temp.left = null;
temp.right = null;
return temp;
}
/* A utility function to insert a new node with
given key in BST */
static Node insert(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.data)
node.left = insert(node.left, key);
else if (key > node.data)
node.right = insert(node.right, key);
/* return the (unchanged) node pointer */
return node;
}
/* Function to count nodes in a binary search tree
using Morris Inorder traversal*/
static int counNodes(Node root)
{
Node current, pre;
// Initialise count of nodes as 0
int count = 0;
if (root == null)
return count;
current = root;
while (current != null)
{
if (current.left == null)
{
// Count node if its left is NULL
count++;
// Move to its right
current = current.right;
}
else
{
/* Find the inorder predecessor of current */
pre = current.left;
while (pre.right != null &&
pre.right != current)
pre = pre.right;
/* Make current as right child of its
inorder predecessor */
if(pre.right == null)
{
pre.right = current;
current = current.left;
}
/* Revert the changes made in if part to
restore the original tree i.e., fix
the right child of predecessor */
else
{
pre.right = null;
// Increment count if the current
// node is to be visited
count++;
current = current.right;
} /* End of if condition pre->right == NULL */
} /* End of if condition current->left == NULL*/
} /* End of while */
return count;
}
/* Function to find median in O(n) time and O(1) space
using Morris Inorder traversal*/
static int findMedian(Node root)
{
if (root == null)
return 0;
int count = counNodes(root);
int currCount = 0;
Node current = root, pre = null, prev = null;
while (current != null)
{
if (current.left == null)
{
// count current node
currCount++;
// check if current node is the median
// Odd case
if (count % 2 != 0 && currCount == (count+1)/2)
return prev.data;
// Even case
else if (count % 2 == 0 && currCount == (count/2)+1)
return (prev.data + current.data)/2;
// Update prev for even no. of nodes
prev = current;
//Move to the right
current = current.right;
}
else
{
/* Find the inorder predecessor of current */
pre = current.left;
while (pre.right != null && pre.right != current)
pre = pre.right;
/* Make current as right child of its inorder predecessor */
if (pre.right == null)
{
pre.right = current;
current = current.left;
}
/* Revert the changes made in if part to restore the original
tree i.e., fix the right child of predecessor */
else
{
pre.right = null;
prev = pre;
// Count current node
currCount++;
// Check if the current node is the median
if (count % 2 != 0 && currCount == (count+1)/2 )
return current.data;
else if (count%2==0 && currCount == (count/2)+1)
return (prev.data+current.data)/2;
// update prev node for the case of even
// no. of nodes
prev = current;
current = current.right;
} /* End of if condition pre->right == NULL */
} /* End of if condition current->left == NULL*/
} /* End of while */
return -1;
}
/* Driver code*/
public static void main(String[] args)
{
/* Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80 */
Node root = null;
root = insert(root, 50);
insert(root, 30);
insert(root, 20);
insert(root, 40);
insert(root, 70);
insert(root, 60);
insert(root, 80);
System.out.println("Median of BST is " + findMedian(root));
}
}
// This code is contributed by prerna saini.
Python 3
# Python program to find closest
# value in Binary search Tree
_MIN=-2147483648
_MAX=2147483648
# Helper function that allocates
# a new node with the given data
# and None left and right poers.
class newNode:
# Constructor to create a new node
def __init__(self, data):
self.data = data
self.left = None
self.right = None
""" A utility function to insert a new node with
given key in BST """
def insert(node,key):
""" If the tree is empty, return a new node """
if (node == None):
return newNode(key)
""" Otherwise, recur down the tree """
if (key < node.data):
node.left = insert(node.left, key)
elif (key > node.data):
node.right = insert(node.right, key)
""" return the (unchanged) node pointer """
return node
""" Function to count nodes in
a binary search tree using
Morris Inorder traversal"""
def counNodes(root):
# Initialise count of nodes as 0
count = 0
if (root == None):
return count
current = root
while (current != None):
if (current.left == None):
# Count node if its left is None
count+=1
# Move to its right
current = current.right
else:
""" Find the inorder predecessor of current """
pre = current.left
while (pre.right != None and
pre.right != current):
pre = pre.right
""" Make current as right child of its
inorder predecessor """
if(pre.right == None):
pre.right = current
current = current.left
else:
pre.right = None
# Increment count if the current
# node is to be visited
count += 1
current = current.right
return count
""" Function to find median in
O(n) time and O(1) space
using Morris Inorder traversal"""
def findMedian(root):
if (root == None):
return 0
count = counNodes(root)
currCount = 0
current = root
while (current != None):
if (current.left == None):
# count current node
currCount += 1
# check if current node is the median
# Odd case
if (count % 2 != 0 and
currCount == (count + 1)//2):
return prev.data
# Even case
elif (count % 2 == 0 and
currCount == (count//2)+1):
return (prev.data + current.data)//2
# Update prev for even no. of nodes
prev = current
#Move to the right
current = current.right
else:
""" Find the inorder predecessor of current """
pre = current.left
while (pre.right != None and
pre.right != current):
pre = pre.right
""" Make current as right child
of its inorder predecessor """
if (pre.right == None):
pre.right = current
current = current.left
else:
pre.right = None
prev = pre
# Count current node
currCount+= 1
# Check if the current node is the median
if (count % 2 != 0 and
currCount == (count + 1) // 2 ):
return current.data
elif (count%2 == 0 and
currCount == (count // 2) + 1):
return (prev.data+current.data)//2
# update prev node for the case of even
# no. of nodes
prev = current
current = current.right
# Driver Code
if __name__ == '__main__':
""" Constructed binary tree is
50
/ \
30 70
/ \ / \
20 40 60 80 """
root = newNode(50)
insert(root, 30)
insert(root, 20)
insert(root, 40)
insert(root, 70)
insert(root, 60)
insert(root, 80)
print("Median of BST is ",findMedian(root))
# This code is contributed
# Shubham Singh(SHUBHAMSINGH10)
C
/* C# program to find the median of BST in O(n)
time and O(1) space*/
using System;
class GfG
{
/* A binary search tree Node has data, pointer
to left child and a pointer to right child */
public class Node
{
public int data;
public Node left, right;
}
// A utility function to create a new BST node
static Node newNode(int item)
{
Node temp = new Node();
temp.data = item;
temp.left = null;
temp.right = null;
return temp;
}
/* A utility function to insert a new node with
given key in BST */
static Node insert(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.data)
node.left = insert(node.left, key);
else if (key > node.data)
node.right = insert(node.right, key);
/* return the (unchanged) node pointer */
return node;
}
/* Function to count nodes in a binary search tree
using Morris Inorder traversal*/
static int counNodes(Node root)
{
Node current, pre;
// Initialise count of nodes as 0
int count = 0;
if (root == null)
return count;
current = root;
while (current != null)
{
if (current.left == null)
{
// Count node if its left is NULL
count++;
// Move to its right
current = current.right;
}
else
{
/* Find the inorder predecessor of current */
pre = current.left;
while (pre.right != null &&
pre.right != current)
pre = pre.right;
/* Make current as right child of its
inorder predecessor */
if(pre.right == null)
{
pre.right = current;
current = current.left;
}
/* Revert the changes made in if part to
restore the original tree i.e., fix
the right child of predecessor */
else
{
pre.right = null;
// Increment count if the current
// node is to be visited
count++;
current = current.right;
} /* End of if condition pre->right == NULL */
} /* End of if condition current->left == NULL*/
} /* End of while */
return count;
}
/* Function to find median in O(n) time and O(1) space
using Morris Inorder traversal*/
static int findMedian(Node root)
{
if (root == null)
return 0;
int count = counNodes(root);
int currCount = 0;
Node current = root, pre = null, prev = null;
while (current != null)
{
if (current.left == null)
{
// count current node
currCount++;
// check if current node is the median
// Odd case
if (count % 2 != 0 && currCount == (count+1)/2)
return prev.data;
// Even case
else if (count % 2 == 0 && currCount == (count/2)+1)
return (prev.data + current.data)/2;
// Update prev for even no. of nodes
prev = current;
//Move to the right
current = current.right;
}
else
{
/* Find the inorder predecessor of current */
pre = current.left;
while (pre.right != null && pre.right != current)
pre = pre.right;
/* Make current as right child of its inorder predecessor */
if (pre.right == null)
{
pre.right = current;
current = current.left;
}
/* Revert the changes made in if part to restore the original
tree i.e., fix the right child of predecessor */
else
{
pre.right = null;
prev = pre;
// Count current node
currCount++;
// Check if the current node is the median
if (count % 2 != 0 && currCount == (count+1)/2 )
return current.data;
else if (count % 2 == 0 && currCount == (count/2)+1)
return (prev.data + current.data)/2;
// update prev node for the case of even
// no. of nodes
prev = current;
current = current.right;
} /* End of if condition pre->right == NULL */
} /* End of if condition current->left == NULL*/
} /* End of while */
return -1;
}
/* Driver code*/
public static void Main(String []args)
{
/* Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80 */
Node root = null;
root = insert(root, 50);
insert(root, 30);
insert(root, 20);
insert(root, 40);
insert(root, 70);
insert(root, 60);
insert(root, 80);
Console.WriteLine("Median of BST is " + findMedian(root));
}
}
// This code is contributed by Arnab Kundu
java 描述语言
<script>
/* JavaScript program to find
the median of BST in O(n)
time and O(1) space*/
/* A binary search tree Node has data, pointer
to left child and a pointer to right child */
class Node {
constructor() {
this.data = 0;
this.left = null;
this.right = null;
}
}
// A utility function to create a new BST node
function newNode(item)
{
var temp = new Node();
temp.data = item;
temp.left = null;
temp.right = null;
return temp;
}
/* A utility function to insert a new node with
given key in BST */
function insert(node , key)
{
/* If the tree is empty, return a new node */
if (node == null) return newNode(key);
/* Otherwise, recur down the tree */
if (key < node.data)
node.left = insert(node.left, key);
else if (key > node.data)
node.right = insert(node.right, key);
/* return the (unchanged) node pointer */
return node;
}
/* Function to count nodes in a binary search tree
using Morris Inorder traversal*/
function counNodes(root)
{
var current, pre;
// Initialise count of nodes as 0
var count = 0;
if (root == null)
return count;
current = root;
while (current != null)
{
if (current.left == null)
{
// Count node if its left is NULL
count++;
// Move to its right
current = current.right;
}
else
{
/* Find the inorder predecessor of current */
pre = current.left;
while (pre.right != null &&
pre.right != current)
pre = pre.right;
/* Make current as right child of its
inorder predecessor */
if(pre.right == null)
{
pre.right = current;
current = current.left;
}
/* Revert the changes made in if part to
restore the original tree i.e., fix
the right child of predecessor */
else
{
pre.right = null;
// Increment count if the current
// node is to be visited
count++;
current = current.right;
} /* End of if condition pre->right == NULL */
} /* End of if condition current->left == NULL*/
} /* End of while */
return count;
} /* Function to find median in O(n) time and O(1) space
using Morris Inorder traversal*/
function findMedian(root)
{
if (root == null)
return 0;
var count = counNodes(root);
var currCount = 0;
var current = root, pre = null, prev = null;
while (current != null)
{
if (current.left == null)
{
// count current node
currCount++;
// check if current node is the median
// Odd case
if (count % 2 != 0 &&
currCount == (count+1)/2)
return prev.data;
// Even case
else if (count % 2 == 0 &&
currCount == (count/2)+1)
return (prev.data + current.data)/2;
// Update prev for even no. of nodes
prev = current;
//Move to the right
current = current.right;
}
else
{
/* Find the inorder predecessor of current */
pre = current.left;
while (pre.right != null &&
pre.right != current)
pre = pre.right;
/* Make current as right child of its
inorder predecessor */
if (pre.right == null)
{
pre.right = current;
current = current.left;
}
/* Revert the changes made in if
part to restore the original
tree i.e., fix the right child of predecessor */
else
{
pre.right = null;
prev = pre;
// Count current node
currCount++;
// Check if the current node is the median
if (count % 2 != 0 &&
currCount == (count+1)/2 )
return current.data;
else if (count%2==0 &&
currCount == (count/2)+1)
return (prev.data+current.data)/2;
// update prev node for the case of even
// no. of nodes
prev = current;
current = current.right;
} /* End of if condition pre->right == NULL */
} /* End of if condition current->left == NULL*/
} /* End of while */
return -1;
}
/* Driver code*/
/* Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80 */
var root = null;
root = insert(root, 50);
insert(root, 30);
insert(root, 20);
insert(root, 40);
insert(root, 70);
insert(root, 60);
insert(root, 80);
document.write("Median of BST is " + findMedian(root));
// This code is contributed by todaysgaurav
</script>
输出:
Median of BST is 50
参考: 【https://www.careercup.com/question?id=4882624968392704】 本文由 萨哈布拉 供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以用write.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 review-team@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
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