Poj 2255 Tree Recovery(二叉搜索树)

题目链接：http://poj.org/problem?id=2255

思路分析：根据先序遍历（如DBACEGF）可以找出根结点(D)，其后为左右子树；根据中序遍历（如ABCDEFG），已知根结点(D),

可以知道在根结点左边的为左子树结点（ABC），右边为右子树结点（EFG）；可以求出左子树与右子树结点个数；已知道左右子树

结点个数（分别为3个），根据先序遍历（如DBACEGF）可知其左子树的先序遍历（BAC）与右子树的先序遍历（EGF）；左右子树

的先序遍历与中序遍历可知，递归构造树再后序遍历求解。

代码如下：

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

struct TreeNode;
typedef TreeNode *SearchTree;
typedef char ElementType;
struct TreeNode
{
    ElementType Element;
    SearchTree Left;
    SearchTree Right;
};

void PostOrder(SearchTree T);
SearchTree Insert(ElementType X, SearchTree T);
SearchTree CreateTree(string PreOrder, string InOrder, SearchTree T);

int main()
{
    string PreOrder, InOrder;

    while (cin >> PreOrder >> InOrder)
    {
        SearchTree T = NULL;

        T = CreateTree(PreOrder, InOrder, T);
        PostOrder(T);
        cout << endl;
    }

    return 0;
}

SearchTree Insert(ElementType X, SearchTree T)
{
    if (T == NULL)
    {
        T = (SearchTree)malloc(sizeof(struct TreeNode));
        if (T == NULL)
        {
            printf("Out of space");
            return NULL;
        }
        T->Element = X;
        T->Left = T->Right = NULL;
    }
    else
    if (X < T->Element)
        T->Left = Insert(X, T->Left);
    else
    if (X > T->Element)
        T->Right = Insert(X, T->Right);

    return T;
}

void PostOrder(SearchTree T)
{
    if (T != NULL)
    {
        PostOrder(T->Left);
        PostOrder(T->Right);
        printf("%c", T->Element);
    }
}

SearchTree CreateTree(string PreOrder, string InOrder, SearchTree T)
{
    int Index;
    char Root;

    if (PreOrder.length() > 0)
    {
        Root = PreOrder[0];
        T = Insert(Root, T);

        Index = InOrder.find(Root);
        T->Left = CreateTree(PreOrder.substr(1, Index), InOrder.substr(0, Index), T->Left);
        T->Right = CreateTree(PreOrder.substr(Index + 1), InOrder.substr(Index + 1), T->Right);
    }

    return T;
}