Print path between any two nodes in a Binary Tree

Given a Binary Tree of distinct nodes and a pair of nodes. The task is to find and print the path between the two given nodes in the binary tree.

For Example, in the above binary tree the path between the nodes 7 and 4 is 7 -> 3 -> 1 -> 4. 

The idea is to find paths from root nodes to the two nodes and store them in two separate vectors or arrays say path1 and path2.

Now, there arises two different cases: 

1. If the two nodes are in different subtrees of root nodes. That is one in the left subtree and the other in the right subtree. In this case it is clear that root node will lie in between the path from node1 to node2. So, print path1 in reverse order and then path 2.
2. If the nodes are in the same subtree. That is either in the left subtree or in the right subtree. In this case you need to observe that path from root to the two nodes will have an intersection point before which the path is common for the two nodes from the root node. Find that intersection point and print nodes from that point in a similar fashion of the above case.

Below is the implementation of the above approach: 

C++

// C++ program to print path between any
// two nodes in a Binary Tree
#include <bits/stdc++.h>
using namespace std;

// structure of a node of binary tree
struct Node {
    int data;
    Node *left, *right;
};

/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
struct Node* getNode(int data)
{
    struct Node* newNode = new Node;
    newNode->data = data;
    newNode->left = newNode->right = NULL;
    return newNode;
}

// Function to check if there is a path from root
// to the given node. It also populates
// 'arr' with the given path
bool getPath(Node* root, vector<int>& arr, int x)
{
    // if root is NULL
    // there is no path
    if (!root)
        return false;

    // push the node's value in 'arr'
    arr.push_back(root->data);

    // if it is the required node
    // return true
    if (root->data == x)
        return true;

    // else check whether the required node lies
    // in the left subtree or right subtree of
    // the current node
    if (getPath(root->left, arr, x) || getPath(root->right, arr, x))
        return true;

    // required node does not lie either in the
    // left or right subtree of the current node
    // Thus, remove current node's value from
    // 'arr'and then return false
    arr.pop_back();
    return false;
}

// Function to print the path between
// any two nodes in a binary tree
void printPathBetweenNodes(Node* root, int n1, int n2)
{
    // vector to store the path of
    // first node n1 from root
    vector<int> path1;

    // vector to store the path of
    // second node n2 from root
    vector<int> path2;

    getPath(root, path1, n1);
    getPath(root, path2, n2);

    int intersection = -1;

    // Get intersection point
    int i = 0, j = 0;
    while (i != path1.size() || j != path2.size()) {

        // Keep moving forward until no intersection
        // is found
        if (i == j && path1[i] == path2[j]) {
            i++;
            j++;
        }
        else {
            intersection = j - 1;
            break;
        }
    }

    // Print the required path
    for (int i = path1.size() - 1; i > intersection; i--)
        cout << path1[i] << " ";

    for (int i = intersection; i < path2.size(); i++)
        cout << path2[i] << " ";
}

// Driver program
int main()
{
    // binary tree formation
    struct Node* root = getNode(0);
    root->left = getNode(1);
    root->left->left = getNode(3);
    root->left->left->left = getNode(7);
    root->left->right = getNode(4);
    root->left->right->left = getNode(8);
    root->left->right->right = getNode(9);
    root->right = getNode(2);
    root->right->left = getNode(5);
    root->right->right = getNode(6);

    int node1 = 7;
    int node2 = 4;
    printPathBetweenNodes(root, node1, node2);

    return 0;
}

Java

// Java program to print path between any
// two nodes in a Binary Tree
import java.util.*;
class Solution
{

// structure of a node of binary tree
static class Node {
    int data;
    Node left, right;
}

/* Helper function that allocates a new node with the
given data and null left and right pointers. */
 static Node getNode(int data)
{
     Node newNode = new Node();
    newNode.data = data;
    newNode.left = newNode.right = null;
    return newNode;
}

// Function to check if there is a path from root
// to the given node. It also populates
// 'arr' with the given path
static boolean getPath(Node root, Vector<Integer> arr, int x)
{
    // if root is null
    // there is no path
    if (root==null)
        return false;

    // push the node's value in 'arr'
    arr.add(root.data);

    // if it is the required node
    // return true
    if (root.data == x)
        return true;

    // else check whether the required node lies
    // in the left subtree or right subtree of
    // the current node
    if (getPath(root.left, arr, x) || getPath(root.right, arr, x))
        return true;

    // required node does not lie either in the
    // left or right subtree of the current node
    // Thus, remove current node's value from
    // 'arr'and then return false
    arr.remove(arr.size()-1);
    return false;
}

// Function to print the path between
// any two nodes in a binary tree
static void printPathBetweenNodes(Node root, int n1, int n2)
{
    // vector to store the path of
    // first node n1 from root
    Vector<Integer> path1= new Vector<Integer>();

    // vector to store the path of
    // second node n2 from root
    Vector<Integer> path2=new Vector<Integer>();

    getPath(root, path1, n1);
    getPath(root, path2, n2);

    int intersection = -1;

    // Get intersection point
    int i = 0, j = 0;
    while (i != path1.size() || j != path2.size()) {

        // Keep moving forward until no intersection
        // is found
        if (i == j && path1.get(i) == path2.get(i)) {
            i++;
            j++;
        }
        else {
            intersection = j - 1;
            break;
        }
    }

    // Print the required path
    for ( i = path1.size() - 1; i > intersection; i--)
        System.out.print( path1.get(i) + " ");

    for ( i = intersection; i < path2.size(); i++)
        System.out.print( path2.get(i) + " ");
}

// Driver program
public static void main(String[] args)
{
    // binary tree formation
     Node root = getNode(0);
    root.left = getNode(1);
    root.left.left = getNode(3);
    root.left.left.left = getNode(7);
    root.left.right = getNode(4);
    root.left.right.left = getNode(8);
    root.left.right.right = getNode(9);
    root.right = getNode(2);
    root.right.left = getNode(5);
    root.right.right = getNode(6);

    int node1 = 7;
    int node2 = 4;
    printPathBetweenNodes(root, node1, node2);

}
}
// This code is contributed by Arnab Kundu

Python

# Python3 program to print path between any
# two nodes in a Binary Tree

import sys
import math

# structure of a node of binary tree
class Node:
    def __init__(self,data):
        self.data = data
        self.left = None
        self.right = None

# Helper function that allocates a new node with the
#given data and NULL left and right pointers.
def getNode(data):
        return Node(data)

# Function to check if there is a path from root
# to the given node. It also populates
# 'arr' with the given path
def getPath(root, rarr, x):

    # if root is NULL
    # there is no path
    if not root:
        return False

    # push the node's value in 'arr'
    rarr.append(root.data)

    # if it is the required node
    # return true
    if root.data == x:
        return True

    # else check whether the required node lies
    # in the left subtree or right subtree of
    # the current node
    if getPath(root.left, rarr, x) or getPath(root.right, rarr, x):
        return True

    # required node does not lie either in the
    # left or right subtree of the current node
    # Thus, remove current node's value from
    # 'arr'and then return false
    rarr.pop()
    return False

# Function to print the path between
# any two nodes in a binary tree
def printPathBetweenNodes(root, n1, n2):

    # vector to store the path of
    # first node n1 from root
    path1 = []

    # vector to store the path of
    # second node n2 from root
    path2 = []
    getPath(root, path1, n1)
    getPath(root, path2, n2)

    # Get intersection point
    i, j = 0, 0
    intersection=-1
    while(i != len(path1) or j != len(path2)):

        # Keep moving forward until no intersection
        # is found
        if (i == j and path1[i] == path2[j]):
            i += 1
            j += 1
        else:
            intersection = j - 1
            break

    # Print the required path
    for i in range(len(path1) - 1, intersection - 1, -1):
        print("{} ".format(path1[i]), end = "")
    for j in range(intersection + 1, len(path2)):
        print("{} ".format(path2[j]), end = "")

# Driver program
if __name__=='__main__':

    # binary tree formation
    root = getNode(0)
    root.left = getNode(1)
    root.left.left = getNode(3)
    root.left.left.left = getNode(7)
    root.left.right = getNode(4)
    root.left.right.left = getNode(8)
    root.left.right.right = getNode(9)
    root.right = getNode(2)
    root.right.left = getNode(5)
    root.right.right = getNode(6)
    node1=7
    node2=4
    printPathBetweenNodes(root,node1,node2)

# This Code is Contributed By Vikash Kumar 37

C#

// C# program to print path between any
// two nodes in a Binary Tree
using System;
using System.Collections.Generic;

class Solution
{

// structure of a node of binary tree
public class Node
{
    public int data;
    public Node left, right;
}

/* Helper function that allocates a new node with the
given data and null left and right pointers. */
static Node getNode(int data)
{
    Node newNode = new Node();
    newNode.data = data;
    newNode.left = newNode.right = null;
    return newNode;
}

// Function to check if there is a path from root
// to the given node. It also populates
// 'arr' with the given path
static Boolean getPath(Node root, List<int> arr, int x)
{
    // if root is null
    // there is no path
    if (root == null)
        return false;

    // push the node's value in 'arr'
    arr.Add(root.data);

    // if it is the required node
    // return true
    if (root.data == x)
        return true;

    // else check whether the required node lies
    // in the left subtree or right subtree of
    // the current node
    if (getPath(root.left, arr, x) || getPath(root.right, arr, x))
        return true;

    // required node does not lie either in the
    // left or right subtree of the current node
    // Thus, remove current node's value from
    // 'arr'and then return false
    arr.RemoveAt(arr.Count-1);
    return false;
}

// Function to print the path between
// any two nodes in a binary tree
static void printPathBetweenNodes(Node root, int n1, int n2)
{
    // vector to store the path of
    // first node n1 from root
    List<int> path1 = new List<int>();

    // vector to store the path of
    // second node n2 from root
    List<int> path2 = new List<int>();

    getPath(root, path1, n1);
    getPath(root, path2, n2);

    int intersection = -1;

    // Get intersection point
    int i = 0, j = 0;
    while (i != path1.Count || j != path2.Count)
    {

        // Keep moving forward until no intersection
        // is found
        if (i == j && path1[i] == path2[i])
        {
            i++;
            j++;
        }
        else
        {
            intersection = j - 1;
            break;
        }
    }

    // Print the required path
    for ( i = path1.Count - 1; i > intersection; i--)
        Console.Write( path1[i] + " ");

    for ( i = intersection; i < path2.Count; i++)
        Console.Write( path2[i] + " ");
}

// Driver code
public static void Main(String[] args)
{
    // binary tree formation
    Node root = getNode(0);
    root.left = getNode(1);
    root.left.left = getNode(3);
    root.left.left.left = getNode(7);
    root.left.right = getNode(4);
    root.left.right.left = getNode(8);
    root.left.right.right = getNode(9);
    root.right = getNode(2);
    root.right.left = getNode(5);
    root.right.right = getNode(6);

    int node1 = 7;
    int node2 = 4;
    printPathBetweenNodes(root, node1, node2);

}
}

// This code is contributed by Princi Singh

Javascript

<script>

    // JavaScript program to print path between any
    // two nodes in a Binary Tree

    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }

    /* Helper function that allocates a new node with the
    given data and null left and right pointers. */
    function getNode(data)
    {
        let newNode = new Node(data);
        return newNode;
    }

    // Function to check if there is a path from root
    // to the given node. It also populates
    // 'arr' with the given path
    function getPath(root, arr, x)
    {
        // if root is null
        // there is no path
        if (root==null)
            return false;

        // push the node's value in 'arr'
        arr.push(root.data);

        // if it is the required node
        // return true
        if (root.data == x)
            return true;

        // else check whether the required node lies
        // in the left subtree or right subtree of
        // the current node
        if (getPath(root.left, arr, x) || getPath(root.right, arr, x))
            return true;

        // required node does not lie either in the
        // left or right subtree of the current node
        // Thus, remove current node's value from
        // 'arr'and then return false
        arr.pop();
        return false;
    }

    // Function to print the path between
    // any two nodes in a binary tree
    function printPathBetweenNodes(root, n1, n2)
    {
        // vector to store the path of
        // first node n1 from root
        let path1= [];

        // vector to store the path of
        // second node n2 from root
        let path2 = [];

        getPath(root, path1, n1);
        getPath(root, path2, n2);

        let intersection = -1;

        // Get intersection point
        let i = 0, j = 0;
        while (i != path1.length || j != path2.length) {

            // Keep moving forward until no intersection
            // is found
            if (i == j && path1[i] == path2[i]) {
                i++;
                j++;
            }
            else {
                intersection = j - 1;
                break;
            }
        }

        // Print the required path
        for ( i = path1.length - 1; i > intersection; i--)
            document.write( path1[i] + " ");

        for ( i = intersection; i < path2.length; i++)
            document.write( path2[i] + " ");
    }

    // binary tree formation
    let root = getNode(0);
    root.left = getNode(1);
    root.left.left = getNode(3);
    root.left.left.left = getNode(7);
    root.left.right = getNode(4);
    root.left.right.left = getNode(8);
    root.left.right.right = getNode(9);
    root.right = getNode(2);
    root.right.left = getNode(5);
    root.right.right = getNode(6);

    let node1 = 7;
    let node2 = 4;
    printPathBetweenNodes(root, node1, node2);

</script>

Output

7 3 1 4 

Complexity Analysis:

• Time Complexity: O(N), as we are using recursion for traversing the tree. Where N is the number of nodes in the tree.
• Auxiliary Space: O(N), as we are using extra space for storing the paths of the trees. Where N is the number of nodes in the tree.