数据结构—前序和中序遍历的二叉树序列还原二叉树

代码：

#include<stdio.h>
#include<stdlib.h>
typedef struct bstTree {char data;struct bstTree* lchild, *rchild;
}bstTree;
bstTree * reStoreTree(char* pre, int preSt, int preEnd, char *mid, int midSt, int midEnd) {//通过前序和中序二叉树遍历还原树if (preSt > preEnd) {return NULL;//如果找不到左子树了则返回NULL
}char rs = pre[preSt];int j;for (j = midSt; j <= midEnd; j++) {if (rs == mid[j]) {break;//在中序序列中找到当前的根节点}}if (j > midEnd) {printf("中序序列有误!\n");return NULL;//不再进行下面子树的操作}int gapLength = j - midSt;//左右子树跟结点的寻找范围bstTree* root = (bstTree*)malloc(sizeof(bstTree));root->data = rs;root->lchild = reStoreTree(pre, preSt + 1, preSt + gapLength, mid, midSt, j - 1);root->rchild = reStoreTree(pre, preSt + gapLength + 1, preEnd, mid, j + 1, midEnd);return root;
}
void prePrint(bstTree*  BSTTree) {//前序遍历二叉树if (BSTTree) {printf("%c ", BSTTree->data);prePrint(BSTTree->lchild);prePrint(BSTTree->rchild);}
}
int main() {char pre[4] = "ABC";char mid[4] = "BAC";bstTree* root = reStoreTree(pre, 0, 2, mid, 0, 2);printf("前序遍历根据前序和中序构造的二叉树\n");prePrint(root);//前序遍历二叉树printf("\n");system("pause");return 0;
}

测试截图：

时间复杂度O(logn),空间复杂度O(1)

如果存在什么问题，欢迎批评指正！谢谢！