Longest Path with Same Values in a Binary Tree

Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root. The length of path between two nodes is represented by the number of edges between them.

Examples:

Input :
              2
             / \
            7   2
           / \   \
          1   1   2
Output : 2

Input :
              4
             / \
            4   4
           / \   \
          4   9   5
Output : 3

The idea is to recursively traverse given binary tree. We can think of any path (of nodes with the same values) in up to two directions(left and right) from it’s root. Then, for each node, we want to know what is the longest possible length extending in the left and the longest possible length extending in the right directions. The longest length that extends from the node will be 1 + length(node->left) if node->left exists, and has the same value as node. Similarly for the node->right case.

While we are computing lengths, each candidate answer will be the sum of the lengths in both directions from that node. We keep updating these answers and return the maximum one.

C++

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// C++ program to find the length of longest
// path with same values in a binary tree.
#include <bits/stdc++.h>
using namespace std;
  
/* A binary tree node has data, pointer to
left child and a pointer to right child */
struct Node {
  int val;
  struct Node *left, *right;
};
  
/* Function to print the longest path
   of same values */
int length(Node *node, int *ans) {
  if (!node)
    return 0;
  
  // Recursive calls to check for subtrees
  int left = length(node->left, ans);
  int right = length(node->right, ans);
  
  // Variables to store maximum lengths in two directions
  int Leftmax = 0, Rightmax = 0;
  
  // If curr node and it's left child has same value
  if (node->left && node->left->val == node->val) 
    Leftmax += left + 1;  
  
  // If curr node and it's right child has same value
  if (node->right && node->right->val == node->val) 
    Rightmax += right + 1;
    
  *ans = max(*ans, Leftmax + Rightmax);
  return max(Leftmax, Rightmax);
}
  
/* Driver function to find length of
   longest same value path*/
int longestSameValuePath(Node *root) {
  int ans = 0;
  length(root, &ans);
  return ans;
}
  
/* Helper function that allocates a
new node with the given data and
NULL left and right pointers. */
Node *newNode(int data) {
  Node *temp = new Node;
  temp->val = data;
  temp->left = temp->right = NULL;
  return temp;
}
  
// Driver code
int main() {
  /* Let us construct a Binary Tree
        4
       / \
      4   4
     / \   \
    4   9   5 */
  
  Node *root = NULL;
  root = newNode(4);
  root->left = newNode(4);
  root->right = newNode(4);
  root->left->left = newNode(4);
  root->left->right = newNode(9);
  root->right->right = newNode(5);
  cout << longestSameValuePath(root);
  return 0;
}

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Python3

# Python3 program to find the length of longest
# path with same values in a binary tree.

# Helper function that allocates a
# new node with the given data and
# None left and right pointers.
class newNode:
def __init__(self, data):
self.val = data
self.left = self.right = None

# Function to prthe longest path
# of same values
def length(node, ans):
if (not node):
return 0

# Recursive calls to check for subtrees
left = length(node.left, ans)
right = length(node.right, ans)

# Variables to store maximum lengths
# in two directions
Leftmax = 0
Rightmax = 0

# If curr node and it’s left child has same value
if (node.left and node.left.val == node.val):
Leftmax += left + 1

# If curr node and it’s right child has same value
if (node.right and node.right.val == node.val):
Rightmax += right + 1

ans[0] = max(ans[0], Leftmax + Rightmax)
return max(Leftmax, Rightmax)

# Driver function to find length of
# longest same value path
def longestSameValuePath(root):
ans = [0]
length(root, ans)
return ans[0]

# Driver code
if __name__ == ‘__main__’:

# Let us construct a Binary Tree
# 4
# / \
# 4 4
# / \ \
# 4 9 5
root = None
root = newNode(4)
root.left = newNode(4)
root.right = newNode(4)
root.left.left = newNode(4)
root.left.right = newNode(9)
root.right.right = newNode(5)
print(longestSameValuePath(root))

# This code is contributed by PranchalK


Output:

3

Complexity Analysis:

  • Time complexity : O(n), where n is the number of nodes in tree as every node is processed once.
  • Auxiliary Space : O(h), where h is the height of tree as recursion can go upto depth h.


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Improved By : PranchalKatiyar



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