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Binary_Tree_Print_All_Root_To_Leaf_Paths.cpp
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Binary_Tree_Print_All_Root_To_Leaf_Paths.cpp
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/*
Introduction
Given a Binary Tree ,Print all its root to leaf paths
Argument/Return Type
Input of total no.of nodes is taken
Input of key values of nodes of tree are taken in level order form
Incase of a null node , -1 is taken as input
Function prints all its root to leaf paths recursively
*/
// Code / Solution
#include <bits/stdc++.h>
using namespace std;
//Define Node as structure
struct Node
{
int key;
Node* left;
Node* right;
};
// Function to create a node with 'value' as the data stored in it.
// Both the children of this new Node are initially null.
struct Node* newNode(int value)
{
Node* n = new Node;
n->key = value;
n->left = NULL;
n->right = NULL;
return n;
}
// Function to build tree with given input
struct Node* createTree(vector<int>v)
{
int n=v.size();
if(n==0)
return NULL;
vector<struct Node* >a(n);
//Create a vector of individual nodes with given node values
for(int i=0;i<n;i++)
{
//If the data is -1 , create a null node
if(v[i]==-1)
a[i] = NULL;
else
a[i] = newNode(v[i]);
}
//Interlink all created nodes to create a tree
//Use two pointers using int to store indexes
//One to keep track of parent node and one for children nodes
for(int i=0,j=1;j<n;i++)
{
//If the parent node is NULL , advance children pointer twice
if(!a[i])
{
j=j+2;
continue;
}
//Connect the two children nodes to parent node
//First left and then right nodes
a[i]->left = a[j++];
if(j<n)
a[i]->right = a[j++];
}
return a[0];
}
//Utility Function to print given path
void PrintThisPath(vector<int>path)
{
for( int i: path ) cout<<i<<" ";
cout<<endl;
return ;
}
//Utility Function to Generate paths
void GeneratePaths(struct Node* root, vector<int>path)
{
if(root==NULL)
{
//A null node is a leaf node obviously
//Hence print its path
PrintThisPath(path);
return;
}
//If its not NULL node , append this node to the path array
path.push_back(root->key);
// If it is a leaf node , print its path
if(root->left==NULL && root->right==NULL)
{
PrintThisPath(path);
return ;
}
//else continue generating paths
GeneratePaths(root->left,path);
GeneratePaths(root->right,path);
return ;
}
// Function to print all root to leaf paths
void PrintRootToLeafPaths(struct Node* root)
{
vector<int>path;
GeneratePaths(root,path);
}
// Driver code
int main()
{
int n;
cout<<"Enter total no.of nodes of the input Tree ( including NULL nodes ) : ";
cin>>n;
vector<int>v(n);
cout<<"Enter value of each node of the tree in level order ( if a node is NULL , enter -1 ) with spaces"<<endl;
for(int i=0;i<n;i++)
{
cin>>v[i]; //store the input values in a vector
}
//create the tree using input node values
struct Node* root=createTree(v);
//Call the function and print the result
cout<<"All root to leaf paths are "<<endl;
PrintRootToLeafPaths(root);
return 0;
}
/*
Input:
0 <= node->key < 1000000000
if node is NULL , -1 is entered as it's key
Sample Test Case 1
Input Binary Tree :
1
/ \
2 11
/ \ / \
3 5 12 13
/ \ / \ / \ / \
4 NULL 6 7 8 9 NULL 4
Input Format :
Example :
Enter total no.of nodes of the input Tree ( including NULL nodes ) : 15
Enter value of each node of the tree in level order ( if a node is NULL , enter -1 ) with spaces
1 2 11 3 5 12 13 4 -1 6 7 8 9 -1 4
Output Format :
Example : ( Output to the above input example )
All root to leaf paths are
1 2 3 4
1 2 3
1 2 5 6
1 2 5 7
1 11 12 8
1 11 12 9
1 11 13
1 11 13 4
Sample Test Case 2
Input Binary Tree :
12
/ \
9 17
/ \ / \
8 10 NULL 18
Input Format :
Example :
Enter total no.of nodes of the input Tree ( including NULL nodes ) : 7
Enter value of each node of the tree in level order ( if a node is NULL , enter -1 ) with spaces
12 9 17 8 10 -1 18
Output Format :
Example : ( Output to the above input example )
All root to leaf paths are
12 9 8
12 9 10
12 17
12 17 18
Time/Space Complexity
Time Complexity : O(n)
Where n is the no.of nodes
Space Complexity : O(n)
Where n is the no.of nodes
*/