给定一个二叉搜索树，找到它的中间值。 如果节点数为偶数：则中值=（（n/2th node+（n+1）/2th node）/2 如果节点数为奇数：则中值=（n+1）/2个节点。 例如，低于BST的中值为12。
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![图片[1]-求O（n）时间和O（1）空间中BST的中值-yiteyi-C++库](https://www.yiteyi.com/wp-content/uploads/geeks/geeks_BST.gif)
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：谷歌
为了找到中位数，我们需要找到BST的顺序，因为它的顺序是按排序的，然后找到中位数，即。 这个想法是基于 BST中使用O（1）额外空间的第K个最小元素 如果允许我们使用额外的空间，那么这个任务非常简单，但是使用递归和堆栈的有序遍历都使用空间，这在这里是不允许的。所以，解决办法是 莫里斯有序遍历 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 << "Median of BST is " << findMedian(root); return 0; } |
JAVA
/* 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. |
Python3
# 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 |
Javascript
<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 萨希尔·查布拉 .如果你喜欢GeekSforgek，并想贡献自己的力量，你也可以使用 写极客。组织 或者把你的文章寄去评论-team@geeksforgeeks.org.看到你的文章出现在Geeksforgeks主页上，并帮助其他极客。 如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。
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