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;
}