根据遍历序列构建二叉树，并转换成双向链表

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RT,根据前序和中序遍历结果，构建二叉树，在此基础上把该二叉树转换成双向链表

#include <iostream>#include <algorithm>using namespace std;typedef struct TNode{        int val;        struct TNode *lchild;        struct TNode *rchild;}TNode, *BTree;TNode *BuildTree(int pre[], int in[], int len) {        if (len <= 0)                return NULL;        TNode *p = (TNode*)malloc(sizeof(TNode));        p->val = pre[0];        int *index = find(in, in+len, pre[0]);        int dis = index - in;        p->lchild = BuildTree(pre+1, in, dis);        p->rchild = BuildTree(pre+dis+1, in+dis+1, len-dis-1);        return p;}void InOrder(BTree T) {//用来测试        if(T == NULL)                return;        InOrder(T->lchild);        cout<<T->val<<" ";        InOrder(T->rchild);}void PostOrder(BTree T) {//用来测试        if(T == NULL)                return;        PostOrder(T->lchild);        PostOrder(T->rchild);        cout<<T->val<<" ";}void BuildBiList(BTree T, TNode **lastnode) {        if (T == NULL)                return ;        TNode * current = T;        BuildBiList(T->lchild, lastnode);        current->lchild = *lastnode;        if (*lastnode != NULL)                (*lastnode)->rchild = current;        *lastnode = current;        BuildBiList(T->rchild, lastnode);}int main() {        //int pre[] = {1, 2, 4, 3};//测试用例        //int in[] = {4, 2, 1, 3};        int pre[] = {5, 3, 1, 4, 8, 6, 9};        int in[] = {1, 3, 4, 5, 6, 8, 9};        BTree T = BuildTree(pre, in, 7);        InOrder(T);//测试构建结果        cout<<endl;        PostOrder(T);        //转换成双向链表        TNode * p = NULL;        BuildBiList(T, &p);//返回的是尾端节点        //逆向打印        while (p != NULL) {                cout << p->val<< " ";                p = p->lchild;        }}

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