重建二叉树

已知前序遍历和中序遍历的结果，重建这棵二叉树

 

#include <iostream>#include <string>using namespace std;//------------------------------------------------------------------// 前序遍历结果：a b d c e f// 中序遍历结果：d b a e c f//------------------------------------------------------------------const int TREELEN = 6;struct NODE{NODE* pLeft;NODE* pRight;char chValue;};void Rebuild(char* pPreOrder, char* pInOrder, int nTreeLen, NODE** pRoot){ //检查边界条件if (pPreOrder==NULL || pInOrder==NULL){return;}//获得前序遍历的第一个结点，即为ROOTNODE *pTemp = new NODE;pTemp->chValue = *pPreOrder;pTemp->pLeft = NULL;pTemp->pRight = NULL;//节点为空，把当前节点复制到根节点if (nTreeLen == 1){return;}//寻找子树长度char* pOrgInOrder = pInOrder;char* pLeftEnd = pInOrder;int nTempLen = 0;//找到左子树结尾while (*pPreOrder!=*pLeftEnd){if (pPreOrder==NULL || pLeftEnd==NULL){return;}nTempLen++;if (nTempLen > nTreeLen){break;}pLeftEnd++;}//寻找左子树长度int nLeftLen = 0;nLeftLen = (int)(pLeftEnd-pOrgInOrder);//寻找右子树长度int nRightLen = 0;nRightLen = nTreeLen - nLeftLen - 1;//rebuild左子树if (nLeftLen>0){Rebuild(pPreOrder+1, pInOrder, nLeftLen, &((*pRoot)->pLeft));}//rebuild右子树if (nRightLen>0){Rebuild(pPreOrder+nLeftLen+1, pInOrder+nLeftLen+1, nRightLen, &((*pRoot)->pRight));}}int main(){char szPreOrder[TREELEN]={'a', 'b', 'd', 'c', 'e', 'f'};char szInOrder[TREELEN]={'d', 'b', 'a', 'e', 'c', 'f'};NODE* pRoot = NULL;Rebuild(szPreOrder,szInOrder,TREELEN,&pRoot);}