珍爱生命，多用python

一道二叉树重建的问题，用C被折磨得不轻，干脆用python做了，一切清爽了不少。

问题：输入二叉树的前序与中序，输出二叉树的后序

本文章来自博客园，如果在别的地方看可能是本人的博客同步

代码

• 1 C代码
• 2 python 代码

1 C代码

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
const int MAXN = 256;

typedef struct TNode{
  int have_value;
  int v;
  struct TNode* left, *right;
} Node;

Node* root;
char s[MAXN + 10];
int failed;
int n = 0, ans[MAXN];

Node* newnode() {
  Node* u = (Node*) malloc(sizeof(Node));
  if(u != NULL) {
    u->have_value = 0;
    u->left = u->right = NULL;
  }
  return u;
}


void addnode(int v, char* s) {
  int n = strlen(s);
  Node* u = root;
  for(int i = 0; i < n; i++)
    if(s[i] == 'L') {
      if(u->left == NULL) u->left = newnode();
      u = u->left;
    } else if(s[i] == 'R') {
      if(u->right == NULL) u->right = newnode();
      u = u->right;
    }
  if(u->have_value) failed = 1;
  u->v = v;
  u->have_value = 1;
}

void remove_tree(Node* u) {
  if(u == NULL) return;
  remove_tree(u->left);
  remove_tree(u->right);
  free(u);
}


int read_input() {
  failed = 0;
  remove_tree(root);
  root = newnode();
  for(;;) {
    if(scanf("%s", s) != 1) return 0;
    if(!strcmp(s, "()")) break;
    int v;
    sscanf(&s[1], "%d", &v);
    addnode(v, strchr(s, ',')+1);
  }
  return 1;
}


int bfs() {
  int front = 0, rear = 1;
  Node* q[MAXN];
  q[0] = root;
  while(front < rear) {
    Node* u = q[front++];
    if(!u->have_value) return 0;
    ans[n++] = u->v;
    if(u->left != NULL) q[rear++] = u->left;
    if(u->right != NULL) q[rear++] = u->right;
  }
  return 1;
}

int main() {
  while(read_input()) {
    if(!bfs()) failed = 1;
    if(failed) printf("-1\n");
    else {
      for(int i = 0; i < n; i++)
        printf("%d ", ans[i]);
      printf("\n");
    }
  }
  return 0;
}

　　

2 python 代码

ans = []
def show_posorder(perorder, inorder):
    if(len(perorder) == 0 and len(inorder) == 0):
        return 1
    else:
        ans.extend(perorder[0])
        ir = inorder.index(perorder[0])
        show_posorder(perorder[1:ir+1], inorder[0:ir]);
        show_posorder(perorder[ir+1:], inorder[ir+1:]);
        print ans.pop(),

def input():
    perorder = raw_input("input the perorder:")
    inorder = raw_input("input the inorder:")
    show_posorder(perorder, inorder)

　　

Author: samael <samael.65535@gmail.com>

Date: 2011-10-17 20:07:42 CST

HTML generated by org-mode 6.21b in emacs 23