ZOJ-1167-Trees on the Level

题解：

我的解法是用一个类似字典树结构的结构体来表示节点。看到另一种解法是用数组来映射二叉树的，开到14000就过了，但是我觉得是数据水了，因为题中说最多 256个节点，如果256个节点连成链型，除根节点外每个节点都是父节点的右儿子。那么数组要开pow(2, 256)个。可见这种方法是不可行的；

• 类似字典树的二叉树

  Accepted

  1167

  C++

  0

  280

  #include "cstdio"
  #include "cstring"
  #include "cstdlib"
  #include "cctype"
  #include "queue"
  #include "vector"
  using namespace std;
  struct Tree {
      int data;
      Tree* lson;
      Tree* rson;
  } *root;
  char s[300];
  // v记录遍历的结果,ok记录可否构成树
  vector<int> v;
  bool ok;
  // 初始化节点
  Tree* init() {
      Tree* point = (Tree*)malloc(sizeof(Tree));
      point->data = 0;
      point->lson = point->rson = NULL;
      return point;
  }
  // 插入一个节点
  void insert(char* s) {
      int _data = 0, i = 1;
      Tree* point = root;
      while (isdigit(s[i])) {
          _data = _data * 10 + (s[i++] ^ '0');
      }
      i++;
      while (s[i] != ')') {
          if (s[i++] == 'L') {
              if (point->lson == NULL) {
                  point->lson = init();
              }
              point = point->lson;
          } else {
              if (point->rson == NULL) {
                  point->rson = init();
              }
              point = point->rson;
          }
      }
      if (point->data) {
          ok = false;
      }
      point->data = _data;
  }
  // 按层遍历并释放二叉树
  void BFS() {
      queue<Tree*> q;
      q.push(root);
      Tree* point;
      while (!q.empty()) {
          point = q.front();
          q.pop();
          v.push_back(point->data);
          if (!point->data) {
              ok = false;
          }
          if (point->lson) {
              q.push(point->lson);
          }
          if (point->rson) {
              q.push(point->rson);
          }
          free(point);
      }
  }
  int main() {
      while (~scanf("%s", s)) {
          root = init();
          v.clear();
          ok = true;
          while(s[2] != '\0') {
              insert(s);
              scanf("%s", s);
          }
          BFS();
          if (!ok) {
              puts("not complete");
              continue;
          }
          printf("%d", v[0]);
          for (int i = 1; i < v.size(); i++) {
              printf(" %d", v[i]);
          }
          puts("");
      }
      return 0;
  }