查找中位数二叉搜索树 [英] Find median in binary search tree

问题描述

编写函数 T ComputeMedian（）const的，计算树的中值在O（n）的时间执行。假设树是BST但不一定平衡。回想一下，n个数字中值的定义如下：如果n是奇数，中位数为x使得值小于x的数目等于值大于x的数。如果n是偶数，则一加值小于x的数目等于值大于x的数。例如，给定的号码8，7,2，5,9，中位数为7，因为有两个值小于7和大于7的两个值如果再加号3到集，中值变为5。

Write the implementation of the function T ComputeMedian() const that computes the median value in the tree in O(n) time. Assume that the tree is a BST but is not necessarily balanced. Recall that the median of n numbers is defined as follows: If n is odd, the median is x such that the number of values smaller than x is equal to the number of values greater than x. If n is even, then one plus the number of values smaller than x is equal to the number of values greater than x. For example, given the numbers 8, 7, 2, 5, 9, the median is 7, because there are two values smaller than 7 and two values larger than 7. If we add number 3 to the set, the median becomes 5.

下面是类的二叉搜索树节点：

Here is the class of binary search tree node:

template <class T>
class BSTNode
{
public:
BSTNode(T& val, BSTNode* left, BSTNode* right);
~BSTNode();
T GetVal();
BSTNode* GetLeft();
BSTNode* GetRight();

private:
T val;
BSTNode* left;
BSTNode* right;  
BSTNode* parent; //ONLY INSERT IS READY TO UPDATE THIS MEMBER DATA
int depth, height;
friend class BST<T>;
};

二叉搜索树类：

Binary search tree class:

template <class T>
class BST
{
public:
BST();
~BST();

bool Search(T& val);
bool Search(T& val, BSTNode<T>* node);
void Insert(T& val);
bool DeleteNode(T& val);

void BFT(void);
void PreorderDFT(void);
void PreorderDFT(BSTNode<T>* node);
void PostorderDFT(BSTNode<T>* node);
void InorderDFT(BSTNode<T>* node);
void ComputeNodeDepths(void);
void ComputeNodeHeights(void);
bool IsEmpty(void);
void Visit(BSTNode<T>* node);
void Clear(void);

private:
BSTNode<T> *root;
int depth;
int count;
BSTNode<T> *med; // I've added this member data.

void DelSingle(BSTNode<T>*& ptr);
void DelDoubleByCopying(BSTNode<T>* node);
void ComputeDepth(BSTNode<T>* node, BSTNode<T>* parent);
void ComputeHeight(BSTNode<T>* node);
void Clear(BSTNode<T>* node);

};

我知道我应该先算树的节点，然后做一个序遍历，直到我到了（N / 2）个节点，并将其返回。我只是不知道如何。

I know I should count the nodes of the tree first and then do an inorder traversal until I reach (n/2)th node and return it. I just have no clue how.

推荐答案

正如你提到的，这是很容易先找到节点的数量，做任何遍历：

As you mentioned, it is fairly easy to first find the number of nodes, doing any traversal:

findNumNodes(node):
   if node == null:
       return 0
   return findNumNodes(node.left) + findNumNodes(node.right) + 1

然后，一个序遍历的终止当节点数量为n / 2

Then, with an inorder traversal that aborts when the node number is n/2:

index=0 //id is a global variable / class variable, or any other variable that is constant between all calls
findMedian(node):
   if node == null:
       return null
   cand = findMedian(node.left)
   if cand != null:
        return cand
   if index == n/2:
       return node
   index = index + 1
   return findMedian(node.right)

的想法是，在序遍历处理节点中的BST按排序方式。因此，自树是BST，在我日您处理节点，是我次才能节点，这当然也适用于我== N / 2 ，当你发现它是 N / 2 个节点，你回吧。

The idea is that in-order traversal processes nodes in BST in sorted manner. So, since the tree is a BST, the ith node you process, is the ith node in order, this is of course also true for i==n/2, and when you find it is the n/2th node, you return it.

作为一个方面说明，您可以添加功能，BST找到我个元素有效（ 0（H），其中 ^ h 是树的高度），使用订单统计树木的。

As a side note, you can add functionality to BST to find ith element efficiently (O(h), where h is the tree's height), using order statistics trees.

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