PAT (Advanced Level) Practise 1099 Build A Binary Search Tree (30)

1099. Build A Binary Search Tree (30)

時間限制

100 ms

記憶體限制

65536 kB

代碼長度限制

16000 B

判題程式

Standard

作者

CHEN, Yue

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
• Given the structure of a binary tree and a sequence of distinct integer keys, there is only one way to fill these keys into the tree so that the resulting tree satisfies the definition of a BST. You are supposed to output the level order traversal sequence of that tree. The sample is illustrated by Figure 1 and 2.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=100) which is the total number of nodes in the tree. The next N lines each contains the left and the right children of a node in the format "left_index right_index", provided that the nodes are numbered from 0 to N-1, and 0 is always the root. If one child is missing, then -1 will represent the NULL child pointer. Finally N distinct integer keys are given in the last line.

Output Specification:

• Sample Input:

• Sample Output:

• 按照排序二叉樹的中序周遊把序列從小到大插入，然後bfs輸出。

#include<cstdio>
#include<vector>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 10;
int ch[maxn][2], n, v[maxn], a[maxn], t;

void dfs(int x)
{
  if (x == -1) return;
  dfs(ch[x][0]);
  v[x] = a[t++];
  dfs(ch[x][1]);
}

int main()
{
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d%d", &ch[i][0], &ch[i][1]);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  sort(a, a + n);
  dfs(0);
  queue<int> p;
  p.push(0);
  while (!p.empty())
  {
    int q = p.front(); p.pop();
    printf("%s%d", q ? " " : "", v[q]);
    if (ch[q][0] >= 0) p.push(ch[q][0]);
    if (ch[q][1] >= 0) p.push(ch[q][1]);
  }
  return 0;
}