Hi Shashikant, Please use this code to get your required solution within time complexity O(n).

ideone.com link : https://ideone.com/wzTO3E

```
/* Constructed binary tree is
6
/ \
3 5
/ \ /
2 5 4
/\
7 4
*/
```

and here is c++ code :

```
#include<bits/stdc++.h>
using namespace std;
struct node
{
int data;
struct node* left;
struct node* right;
};
void printPathsRecur(struct node* node, int path[], int pathLen);
void printArray(int ints[], int len);
void printPaths(struct node* node)
{
int path[1000];
printPathsRecur(node, path, 0);
}
void printPathsRecur(struct node* node, int path[], int pathLen)
{
if (node==NULL)
return;
path[pathLen] = node->data;
pathLen++;
if (node->left==NULL && node->right==NULL)
{
printArray(path, pathLen);
}
else
{
printPathsRecur(node->left, path, pathLen);
printPathsRecur(node->right, path, pathLen);
}
}
void printArray(int ints[], int len)
{
int i;
for (i=0; i<len; i++)
{
printf("%d ", ints[i]);
}
printf("\n");
}
struct node* newnode(int data)
{
struct node* node = (struct node*)
malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
return(node);
}
int main()
{
/* Constructed binary tree is
6
/ \
3 5
/ \ /
2 5 4
/\
7 4
*/
struct node *root = newnode(6);
root->left = newnode(3);
root->right = newnode(5);
root->left->left = newnode(2);
root->left->right = newnode(5);
root->left->right->left=newnode(7);
root->left->right->right=newnode(4);
root->right->left = newnode(4);
printPaths(root);
getchar();
return 0;
}
```