# Shortest distance between two nodes in an infinite binary tree

Consider you have an infinitely long binary tree having a pattern as below:

1 / \ 2 3 / \ / \ 4 5 6 7 / \ / \ / \ / \ . . . . . . . .

Given two nodes with values x and y. The task is to find the length of the shortest path between the two nodes.

**Examples:**

Input: x = 2, y = 3 Output: 2 Input: x = 4, y = 6 Output: 4

A **naive approach** is to store all the ancestors of both nodes in 2 Data-structures(vectors, arrays, etc..) and do a binary search for the first element(let index i) in vector1, and check if it exists in the vector2 or not. If it does, return the index(let x) of the element in vector2. The answer will be thus

distance = v1.size() – 1 – i + v2.size() – 1 – x

Below is the implementation of the above approach.

## C++

`// CPP program to find distance ` `// between two nodes ` `// in a infinite binary tree ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// to stores ancestors of first given node ` `vector<` `int` `> v1; ` `// to stores ancestors of first given node ` `vector<` `int` `> v2; ` ` ` `// normal binary search to find the element ` `int` `BinarySearch(` `int` `x) ` `{ ` ` ` `int` `low = 0; ` ` ` `int` `high = v2.size() - 1; ` ` ` ` ` `while` `(low <= high) { ` ` ` `int` `mid = (low + high) / 2; ` ` ` ` ` `if` `(v2[mid] == x) ` ` ` `return` `mid; ` ` ` `else` `if` `(v2[mid] > x) ` ` ` `high = mid - 1; ` ` ` `else` ` ` `low = mid + 1; ` ` ` `} ` ` ` `return` `-1; ` `} ` ` ` `// function to make ancestors of first node ` `void` `MakeAncestorNode1(` `int` `x) ` `{ ` ` ` `v1.clear(); ` ` ` `while` `(x) { ` ` ` `v1.push_back(x); ` ` ` `x /= 2; ` ` ` `} ` ` ` `reverse(v1.begin(), v1.end()); ` `} ` ` ` `// function to make ancestors of second node ` `void` `MakeAncestorNode2(` `int` `x) ` `{ ` ` ` `v2.clear(); ` ` ` `while` `(x) { ` ` ` `v2.push_back(x); ` ` ` `x /= 2; ` ` ` `} ` ` ` `reverse(v2.begin(), v2.end()); ` `} ` ` ` `// function to find distance bewteen two nodes ` `int` `Distance() ` `{ ` ` ` `for` `(` `int` `i = v1.size() - 1; i >= 0; i--) { ` ` ` `int` `x = BinarySearch(v1[i]); ` ` ` `if` `(x != -1) { ` ` ` `return` `v1.size() - 1 - i + v2.size() - 1 - x; ` ` ` `} ` ` ` `} ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `node1 = 2, node2 = 3; ` ` ` ` ` `// find ancestors ` ` ` `MakeAncestorNode1(node1); ` ` ` `MakeAncestorNode2(node2); ` ` ` ` ` `cout << ` `"Distance between "` `<< node1 << ` ` ` `" and "` `<< node2 << ` `" is : "` `<< Distance(); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

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## Python3

# Python3 program to find the distance between

# two nodes in an infinite binary tree

# normal binary search to find the element

def BinarySearch(x):

low = 0

high = len(v2) – 1

while low <= high:
mid = (low + high) // 2
if v2[mid] == x:
return mid
elif v2[mid] > x:

high = mid – 1

else:

low = mid + 1

return -1

# Function to make ancestors of first node

def MakeAncestorNode1(x):

v1.clear()

while x:

v1.append(x)

x //= 2

v1.reverse()

# Function to make ancestors of second node

def MakeAncestorNode2(x):

v2.clear()

while x:

v2.append(x)

x //= 2

v2.reverse()

# Function to find distance bewteen two nodes

def Distance():

for i in range(len(v1) – 1, -1, -1):

x = BinarySearch(v1[i])

if x != -1:

return (len(v1) – 1 – i +

len(v2) – 1 – x)

# Driver code

if __name__ == “__main__”:

node1, node2 = 2, 3

v1, v2 = [], []

# Find ancestors

MakeAncestorNode1(node1)

MakeAncestorNode2(node2)

print(“Distance between”, node1,

“and”, node2, “is :”, Distance())

# This code is contributed by Rituraj Jain

**Output:**

Distance between 2 and 3 is : 2

**Time Complexity:** O(log(max(x, y)) * log(max(x, y)))

**Auxiliary Space:** O(log(max(x, y)))

An **efficient approach** is to use the property of 2*x and 2*x+1 given. Keep dividing the larger of the two nodes by 2. If the larger becomes the smaller one, then divide the other one. At a stage, both the values will be the same, keep a count on the number of divisions done which will be the answer.

Below is the implementation of the above approach.

## C++

`// C++ program to find the distance ` `// between two nodes in an infinite ` `// binary tree ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// function to find the distance ` `// between two nodes in an infinite ` `// binary tree ` `int` `Distance(` `int` `x, ` `int` `y) ` `{ ` ` ` `// swap the smaller ` ` ` `if` `(x < y) { ` ` ` `swap(x, y); ` ` ` `} ` ` ` `int` `c = 0; ` ` ` ` ` `// divide till x!=y ` ` ` `while` `(x != y) { ` ` ` ` ` `// keep a count ` ` ` `++c; ` ` ` ` ` `// perform divison ` ` ` `if` `(x > y) ` ` ` `x = x >> 1; ` ` ` ` ` `// when the smaller ` ` ` `// becomes the greater ` ` ` `if` `(y > x) { ` ` ` `y = y >> 1; ` ` ` `++c; ` ` ` `} ` ` ` `} ` ` ` `return` `c; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `x = 4, y = 6; ` ` ` `cout << Distance(x, y); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Python3

# Python3 program to find the distance between

# two nodes in an infinite binary tree

# Function to find the distance between

# two nodes in an infinite binary tree

def Distance(x, y):

# Swap the smaller

if x < y:
x, y = y, x
c = 0
# divide till x != y
while x != y:
# keep a count
c += 1
# perform divison
if x > y:

x = x >> 1

# when the smaller becomes

# the greater

if y > x:

y = y >> 1

c += 1

return c

# Driver code

if __name__ == “__main__”:

x, y = 4, 6

print(Distance(x, y))

# This code is contributed by

# Rituraj Jain

**Output:**

4

**Time Complexity:** O(log(max(x, y)))

**Auxiliary Space:** O(1)

The efficient approach has been suggested by Striver.

## Recommended Posts:

- Shortest path between two nodes in array like representation of binary tree
- Find distance between two nodes of a Binary Tree
- Queries to find distance between two nodes of a Binary tree
- Distance between two nodes of binary tree with node values from 1 to N
- Queries to find distance between two nodes of a Binary tree - O(logn) method
- Print the first shortest root to leaf path in a Binary Tree
- Find distance from root to given node in a binary tree
- Sum of all nodes in a binary tree
- Tree with N nodes and K leaves such that distance between farthest leaves is minimized
- Sum of all the Boundary Nodes of a Binary Tree
- Sink Odd nodes in Binary Tree
- Sum of all leaf nodes of binary tree
- XOR of path between any two nodes in a Binary Tree
- Product of all nodes in a Binary Tree
- Sum of nodes in top view of binary tree

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