1099. Build A Binary Search Tree (30)

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.
• Both the left and right subtrees must also be binary search trees.

  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:

  For each test case, print in one line the level order traversal sequence of that tree. All the numbers must be separated by a space, with no extra space at the end of the line.

  Sample Input:

  9
  1 6
  2 3
  -1 -1
  -1 4
  5 -1
  -1 -1
  7 -1
  -1 8
  -1 -1
  73 45 11 58 82 25 67 38 42
  

  Sample Output:

  58 25 82 11 38 67 45 73 42

• #include <iostream>
  #include <cstdio>
  #include <algorithm>
  #include <map>
  #include <queue>
  #include <vector>
  using namespace std;
  const int maxn = 110;
  int tree[maxn][2];
  int node[maxn];
  map<int, int> m;
  vector<int> v;
  queue<int> Q;
  void dfs(int x)
  {
  if(tree[x][0] != -1) 
  dfs(tree[x][0]);
  v.push_back(x);
  if(tree[x][1] != -1)
  dfs(tree[x][1]);
  return;
  }
  void bfs(int x)
  {
  int root;
  while(true){
  if(tree[x][0] != -1)
  Q.push(tree[x][0]);
  if(tree[x][1] != -1)
  Q.push(tree[x][1]);
  root = Q.front();
  printf("%d", m[root]);
  Q.pop();
  if(Q.empty())
  break;
  else{
  printf(" ");
  x = Q.front();
  }
  }
  }
  int main()
  {
  int n;
  scanf("%d", &n);
  for(int i = 0; i < n; i++)
  scanf("%d %d", &tree[i][0], &tree[i][1]);
  for(int i = 0; i < n; i++)
  scanf("%d", &node[i]);
  sort(node, node + n);
  dfs(0);
  for(int i = 0; i < n; i++)
  m[v[i]] = node[i];
  Q.push(0);
  bfs(0);
  return 0;
  }