二叉树系列

引言

本文来自于Google的一道题目：

how to merge two binary search tree into balanced binary search tree. how to merge two binary search tree into balanced binary search tree.. Let there be m elements in first tree and n elements in the other tree. Your merge function should take O(m+n) time and do it in place.

http://www.careercup.com/question?id=5261732222074880

要线性时间内完成，而且要求in place，不难想到把BST转换为Double linked list。

然后将两个dll merge。但是，存在线性时间复杂度的算法，把dll转化成BST吗？

下面就是本文要记述的内容，算法参考自

http://www.geeksforgeeks.org/in-place-conversion-of-sorted-dll-to-balanced-bst/

已排序的 Double linked list 转化为 BST

因为要求线性时间，所以必须保证DLL上的每个节点只被访问 consant 次，最好是一次。

既然要求每个节点只能被访问一次，那么从根节点构造肯定是不行的。这种算法让BST从叶节点开始被构造，通过灵活地运用递归，巧妙地实现了自底向上构造BST的过程，而且还能保证BST是平衡的。

#include<iostream>
#include<stack>
using namespace std;

struct BSTNode{
    int val;
    BSTNode *left;
    BSTNode *right;
    BSTNode(int v): val(v), left(NULL), right(NULL){}
};

class BSTtoDLL{
    public:
        BSTNode *func(BSTNode *root){
            BSTtoDLLCore(root);
            return head;
        }

    private:
        BSTNode* pre = NULL;
        BSTNode* head = NULL;
        void BSTtoDLLCore(BSTNode *root){
            if(!root) return;
            BSTtoDLLCore(root -> left);
            if(pre){
                pre -> right = root;
                root -> left = pre;
            }else{
                head = root;
            }
            pre = root;
            BSTtoDLLCore(root -> right);
        }
};

class DLLtoBalancedBST{
    public:
        BSTNode* func(BSTNode* head){
            if(!head) return head;
            int n = 0;
            for(BSTNode *p = head; p; ++n, p = p -> right);
            return DLLtoBalancedBSTCore(&head, n);
        }
    private:
        //DLL to BST in place, Time O(N), Space O(LogN) for stack, N is the amount of nodes.
        //DLL needs to be sorted.
        BSTNode* DLLtoBalancedBSTCore(BSTNode** headref, int n){
            if(n == 0) return NULL;
            BSTNode* left = DLLtoBalancedBSTCore(headref, n/2);
            BSTNode *root = *headref;
            root -> left = left;
            *headref = root -> right;
            root -> right = DLLtoBalancedBSTCore(headref, n-n/2-1);
            return root;
        }

};

void InorderPrint(BSTNode* root){
    if(!root) return;
    stack<BSTNode *> st;
    while(!st.empty() || root){
        if(!root){
            root = st.top();
            st.pop();
            cout << root -> val << ' ';
            root = root -> right;
        }else{
            st.push(root);
            root = root -> left;
        }
    }
    cout << endl;
}


int main(){
    
    //Construct oringal BST
    BSTNode *root = new BSTNode(5);
    BSTNode *left3 = new BSTNode(3);
    BSTNode *left1 = new BSTNode(1);
    BSTNode *left2 = new BSTNode(2);
    BSTNode *right6 = new BSTNode(6);
    BSTNode *right7 = new BSTNode(7);
    
    root -> left = left2;
    left2 -> left = left1;
    left2 -> right = left3;
    root -> right = right7;
    right7 -> left = right6;
    
    cout << "-------Inorder print BST-------
";
    InorderPrint(root);

    //Convert BST to DLL
    BSTtoDLL bstdll;
    BSTNode *head = bstdll.func(root);
    BSTNode *p = head;
    cout << "-------print converted double linked list----------
";
    for(; p->right != NULL; cout << p -> val << ' ', p = p -> right);
    cout << endl;
    for(; p != NULL; cout << p -> val << ' ', p = p -> left);
    cout << endl;

    //Convert DLL back to Balenced BST
    DLLtoBalancedBST dllbst;
    BSTNode *con_root = dllbst.func(head);
    cout << "-------Inorder print converted BST-------
";
    InorderPrint(con_root);

    return 0;
}

高亮部分为转化过程。

每次递归，headaddr这个指向节点的指针向末尾移动一次。因此每个节点只被访问一次，时间复杂度是线性的。

我们可以发现，这种算法对单向链表也适用。当然单链表不能保证in place，必须新申明节点，但是时间复杂度依然是线性的。

下面给出单向链表上的实现。

已排序的单向Linked list 转化为BST

#include<iostream>
#include<stack>
using namespace std;

struct ListNode{
    int val;
    ListNode* next;
    ListNode(int v): val(v), next(NULL){}
};

struct BSTnode{
    int val;
    BSTnode* left;
    BSTnode* right;
    BSTnode(int v): val(v), left(NULL), right(NULL){}
};

BSTnode *LLtoBSTCore(ListNode **headaddr, int n){
    if(n <= 0) return NULL;
    BSTnode *left = LLtoBSTCore(headaddr, n/2);
    BSTnode *root = new BSTnode((*headaddr) -> val);
    root -> left = left;
    *headaddr = (*headaddr) -> next;
    root -> right = LLtoBSTCore(headaddr, n-n/2-1);
    return root;
}

BSTnode *LLtoBST(ListNode *head){
    if(!head) return NULL;
    int n = 0; ListNode *p = head;
    for(; p; ++n, p = p -> next);
    return LLtoBSTCore(&head, n);
}

int main(){
    ListNode *head = new ListNode(1);
    ListNode *end = head;
    for(int i = 2; i <= 9; end -> next = new ListNode(i++), end = end -> next);
    cout << "List: 
";
    for(ListNode *p = head; p; cout << p -> val << ' ', p = p -> next);
    cout << endl;

    BSTnode *root = LLtoBST(head);

    cout << "BST inorder: " << endl;
    stack<BSTnode *> st;
    while(!st.empty() || root){
        if(!root){
            root = st.top();
            st.pop();
            cout << root -> val << " ";
            root = root -> right;
        }else{
            st.push(root);
            root = root -> left;
        }
    }
    cout << endl;
    
}

高亮部分为转化过程。

给一个已排序的序列的 Iterator，将序列转化为BST

再推而广之，对于给定一个只能向 next 移动的iterator，通过这个算法也能构造将 iterator 经过的节点构造为BST。不过我们事先知道或者计算出序列的长度。

这里给出C#的实现代码。IEnumerator 就是一个迭代器，支持MoveNext() - 将迭代器往后移，Current 属性 - 返回迭代器当前所指向的值。

高亮部分为基于IEnumrator的构造过程。

因为接口IEnumerator 不是.NET CLR 中的 primitive type(基元类型)，因此将 IEnumerator 赋值或者作为函数参数时，都是按引用传递，这正是我们需要的。因此C#的实现代码省去了C++中使用 两个**的麻烦。

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApplication1
{
    public class BSTNode<T>
    {
        public T val;
        public BSTNode<T> left;
        public BSTNode<T> right;
        public BSTNode(T v)
        {
            val = v;
            left = null;
            right = null;
        }
    }

    /// <summary>
    /// Build Binary search tree from given iterator
    /// </summary>
    public class BuildBST
    {
        public BSTNode<T> BuildFromIEt<T>(IEnumerator<T> it, int n)
        {
            if (n == 0) return null;
            BSTNode<T> left = BuildFromIEt(it, n / 2);
            it.MoveNext();
            BSTNode<T> node = new BSTNode<T>(it.Current);
            node.left = left;
            node.right = BuildFromIEt(it, n - n / 2 - 1);

            return node;
        }
    }

    /// <summary>
    /// A class to out put nodes in a tree
    /// </summary>
    public class TreeTrav
    {
        public static void Inorde<T>(BSTNode<T> root)
        {
            if (root == null) return;
            Stack<BSTNode<T>> st = new Stack<BSTNode<T>>();
            while (root != null || st.Count > 0)
            {
                if (root == null)
                {
                    root = st.Peek();
                    st.Pop();
                    Console.Write(root.val.ToString() + ' ');
                    root = root.right;
                }
                else
                {
                    st.Push(root);
                    root = root.left;
                }
            }
            Console.WriteLine();
        }
    }

    class Program
    {

        static void Main(string[] args)
        {
            IEnumerable<int> list = new List<int>() { 1, 2, 3, 6, 7, 9 };

            Console.WriteLine("----Original list----");
            for (IEnumerator<int> it = list.GetEnumerator(); it.MoveNext(); Console.Write(it.Current.ToString() + ' ')) ;
            Console.WriteLine();

            BuildBST buildBST = new BuildBST();
            BSTNode<int> root = buildBST.BuildFromIEt(list.GetEnumerator(), list.Count());
            Console.WriteLine("----Inorder traverse generated BST----");
            TreeTrav.Inorde(root);
            Console.Read();
        }
    }
}