Puzzle: Find the order of n persons standing in a line (based on their heights)

Question

Saw this question on Careercup.com:

Given heights of n persons standing in a line and a list of numbers corresponding to each person (p) that gives the number of pers

Answer 1

I found this kind of problem on SPOJ. I created a binary tree with little variation. When a new height is inserted, if the front is smaller than the root's front then it goes to the left otherwise right.

Here is the C++ implementation:

#include
using namespace std;

struct TreeNode1
{
    int val;
    int _front;
    TreeNode1* left;
    TreeNode1*right;
};

TreeNode1* Add(int x, int v)
{
    TreeNode1* p= (TreeNode1*) malloc(sizeof(TreeNode1));

    p->left=NULL;
    p->right=NULL;
    p->val=x;
    p->_front=v;

    return p;
}

TreeNode1* _insert(TreeNode1* root, int x, int _front)
{
    if(root==NULL) return Add(x,_front);
    if(root->_front >=_front)
    {
            root->left=_insert(root->left,x,_front);
            root->_front+=1;
    }
    else
    {
        root->right=_insert(root->right,x,_front-root->_front);
    }

    return root;
}

bool comp(pair a, pair b)
{
    return a.first>b.first;
}

void in_order(TreeNode1 * root, vector&v)
{
    if(root==NULL) return ;

    in_order(root->left,v);

    v.push_back(root->val);

    in_order(root->right,v);


}

vectorsoln(vectorh, vectorin )
{
    vector >vc;
    for(int i=0;iv;
    in_order(root,v);
    return v;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
            vectorh;
            vectorin;

            for(int i=0;i>x;
            h.push_back(x);}
            for(int i=0;i>x;
            in.push_back(x);}

            vectorv=soln(h,in);
            for(int i=0;i