还原二叉树

来源：互联网发布：hao123网址之家淘宝编辑：程序博客网时间：2024/05/16 08:35

原题：http://www.patest.cn/contests/mooc-ds2015spring/03-%E6%A0%912（Tree Traversals Again）

思路：push结点的顺序就是前序遍历的顺序，pop的顺序就是中序遍历的顺序，跟据前序与中序可还原二叉树

代码：

#include <cstdio>#include <iostream>#include <stack>using namespace std;typedef struct node{int data;struct node *left;struct node *right;}*BinTree;BinTree pre_in(int *pre, int *in, int n){if (n == 0)return NULL;BinTree T = (BinTree)malloc(sizeof(struct node));T->data = *pre;int *p = in;for (; p < in + n; ++p){if (*p == *pre) break;}int k = p - in;T->left = pre_in(pre + 1, in, k);T->right = pre_in(pre + k + 1, in + k + 1, n - k - 1);return T;}bool first = true;   //放到函数外作为全局变量 不然每次递归调用first都是新的各层函数内局部变量 都是truevoid PostOrder(BinTree T){if (T){PostOrder(T->left);PostOrder(T->right);if (first){printf("%d", T->data);first = false;}else printf(" %d", T->data);}}int main(){int n,t;//t存放push后面的数char a[5];  //存放push或popint pre[30],in[30];int i,j; //作为pre和in的索引i = j = 0;stack<int> s;cin >> n;while (j < n){scanf("%s", a);if (a[1] == 'u') {scanf("%d", &t);s.push(t);pre[i++] = t;}else {in[j++] = s.top();s.pop();}}BinTree T = pre_in(pre, in, n);PostOrder(T);return 0;}

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