PAT1099

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
  #include
  #include
  #include 
  using namespace std;
  struct Node
  {
      int val;
      int lChild;
      int rChild;
      Node() :val(0), lChild(-1), rChild(-1){
  
      }
  };
  struct Node node[105];
  int num[105];
  int cnt = 0;
  void inorderTraverse(int root){
      if (root != -1){
          inorderTraverse(node[root].lChild);
          node[root].val = num[cnt++];
          inorderTraverse(node[root].rChild);
      }
  }
  int main()
  {
      int N;
      cin >> N;
      stack s;
      s.push(0);
      int num1, num2,idx;
      for (int i = 0; i < N;i++){
          cin >> num1 >> num2;
          node[i].lChild = num1;
          node[i].rChild = num2;
      }
      for (int i = 0; i < N; i++){
          scanf("%d",&num[i]);
      }
      sort(num,num+N);
      inorderTraverse(0);
      queue q;
      q.push(0);
  	int times=1;
      while (!q.empty()){
          idx = q.front();
          q.pop();
          if(times){
          printf("%d",node[idx].val);
      	times=0;
  		}
  		else {
          	 printf(" %d",node[idx].val);
  		}
          if (node[idx].lChild != -1){
              q.push(node[idx].lChild);
          }
          if (node[idx].rChild != -1){
              q.push(node[idx].rChild);
          }
      }
      return 0;
  }