Given a binary tree, find the path length having maximum number of bends.
Note: Here, bend indicates switching from left to right or vice versa while traversing in the tree.
For example, consider below paths (L means moving leftwards, R means moving rightwards):
LLRRRR – 1 Bend
RLLLRR – 2 Bends
LRLRLR – 5 Bends
Prerequisite : Finding Max path length in a Binary Tree
Examples:
Input :
4
/ \
2 6
/ \ / \
1 3 5 7
/
9
/ \
12 10
\
11
/ \
45 13
\
14
Output : 6
In the above example, the path 4-> 6-> 7-> 9-> 10-> 11-> 45
is having the maximum number of bends, i.e., 3.
The length of this path is 6.
Approach :
The idea is to traverse the tree for left and right subtrees of the root. While traversing, keep track of the direction of motion (left or right). Whenever, direction of motion changes from left to right or vice versa increment the number of bends in the current path by 1.
On reaching the leaf node, compare the number of bends in the current path with the maximum number of bends(i.e., maxBends) seen so far in a root-to-leaf path. If the number of bends in the current path is greater than the maxBends, then update the maxBends equal to the number of bends in the current path and update the maximum path length (i.e., len) also to the length of the current path.
Implementation :
C++
// C++ program to find path length// having maximum number of bends#include <bits/stdc++.h>using namespace std;// structure nodestruct Node { int key; struct Node* left; struct Node* right;};// Utility function to create a new nodestruct Node* newNode(int key){ struct Node* node = new Node(); node->left = NULL; node->right = NULL; node->key = key; return node;}/* Recursive function to calculate the pathlength having maximum number of bends.The following are parameters for this function.node --> pointer to the current nodedir --> determines whether the current nodeis left or right child of it's parent nodebends --> number of bends so far in thecurrent path.maxBends --> maximum number of bends in apath from root to leafsoFar --> length of the current path sofar traversedlen --> length of the path having maximumnumber of bends*/void findMaxBendsUtil(struct Node* node, char dir, int bends, int* maxBends, int soFar, int* len){ // Base Case if (node == NULL) return; // Leaf node if (node->left == NULL && node->right == NULL) { if (bends > *maxBends) { *maxBends = bends; *len = soFar; } } // Recurring for both left and right child else { if (dir == 'l') { findMaxBendsUtil(node->left, dir, bends, maxBends, soFar + 1, len); findMaxBendsUtil(node->right, 'r', bends + 1, maxBends, soFar + 1, len); } else { findMaxBendsUtil(node->right, dir, bends, maxBends, soFar + 1, len); findMaxBendsUtil(node->left, 'l', bends + 1, maxBends, soFar + 1, len); } }}// Helper function to call findMaxBendsUtil()int findMaxBends(struct Node* node){ if (node == NULL) return 0; int len = 0, bends = 0, maxBends = -1; // Call for left subtree of the root if (node->left) findMaxBendsUtil(node->left, 'l', bends, &maxBends, 1, &len); // Call for right subtree of the root if (node->right) findMaxBendsUtil(node->right, 'r', bends, &maxBends, 1, &len); // Include the root node as well in the path length len++; return len;}// Driver codeint main(){ /* Constructed binary tree is 10 / \ 8 2 / \ / 3 5 2 \ 1 / 9 */ struct Node* root = newNode(10); root->left = newNode(8); root->right = newNode(2); root->left->left = newNode(3); root->left->right = newNode(5); root->right->left = newNode(2); root->right->left->right = newNode(1); root->right->left->right->left = newNode(9); cout << findMaxBends(root) - 1; return 0;} |
Java
// Java program to find path length// having maximum number of bendsimport java.util.*;class GFG{ // structure node static class Node { int key; Node left; Node right; }; // Utility function to create a new node static Node newNode(int key) { Node node = new Node(); node.left = null; node.right = null; node.key = key; return node; } /* Recursive function to calculate the pathlength having maximum number of bends.The following are parameters for this function.node -. pointer to the current nodedir -. determines whether the current nodeis left or right child of it's parent nodebends -. number of bends so far in thecurrent path.maxBends -. maximum number of bends in apath from root to leafsoFar -. length of the current path sofar traversedlen -. length of the path having maximumnumber of bends*/ static int maxBends; static int len; private static final char LEFT = 'l'; private static final char RIGHT = 'r'; static void findMaxBendsUtil(Node node, char dir, int bends, int lenSoFar) { // Base Case if (node == null) return; // Leaf node if (node.left == null && node.right == null) { if (bends > maxBends) { maxBends = bends; len = lenSoFar; } } // Recurring for both left and right child else { if (dir == LEFT) { findMaxBendsUtil(node.left, dir, bends, lenSoFar + 1); findMaxBendsUtil(node.right, RIGHT, bends + 1, lenSoFar + 1); } else { findMaxBendsUtil(node.right, dir, bends, lenSoFar + 1); findMaxBendsUtil(node.left, LEFT, bends + 1, lenSoFar + 1); } } } // Helper function to call findMaxBendsUtil() static int findMaxBends(Node node) { if (node == null) return 0; len = 0; maxBends = -1; int bends = 0; // Call for left subtree of the root if (node.left != null) findMaxBendsUtil(node.left, LEFT, bends, 1); // Call for right subtree of the root if (node.right != null) findMaxBendsUtil(node.right, RIGHT, bends, 1); // Include the root node as well in the path length len++; return len; } // Driver code public static void main(String[] args) { /* Constructed binary tree is 10 / \ 8 2 / \ / 3 5 2 \ 1 / 9 */ Node root = newNode(10); root.left = newNode(8); root.right = newNode(2); root.left.left = newNode(3); root.left.right = newNode(5); root.right.left = newNode(2); root.right.left.right = newNode(1); root.right.left.right.left = newNode(9); System.out.print(findMaxBends(root) - 1); }}// This code is contributed by Rajput-Ji |
Python3
# Python3 program to find path Length # having maximum number of bends # Utility function to create a new node class newNode: def __init__(self, key): self.left = None self.right = None self.key = key # Recursive function to calculate the path # Length having maximum number of bends. # The following are parameters for this function. # node -. pointer to the current node # Dir -. determines whether the current node # is left or right child of it's parent node # bends -. number of bends so far in the # current path. # maxBends -. maximum number of bends in a # path from root to leaf # soFar -. Length of the current path so # far traversed # Len -. Length of the path having maximum # number of bends def findMaxBendsUtil(node, Dir, bends, maxBends, soFar, Len): # Base Case if (node == None): return # Leaf node if (node.left == None and node.right == None): if (bends > maxBends[0]): maxBends[0] = bends Len[0] = soFar # Having both left and right child else: if (Dir == 'l'): findMaxBendsUtil(node.left, Dir, bends, maxBends, soFar + 1, Len) findMaxBendsUtil(node.right, 'r', bends + 1, maxBends, soFar + 1, Len) else: findMaxBendsUtil(node.right, Dir, bends, maxBends, soFar + 1, Len) findMaxBendsUtil(node.left, 'l', bends + 1, maxBends, soFar + 1, Len)# Helper function to call findMaxBendsUtil() def findMaxBends(node): if (node == None): return 0 Len = [0] bends = 0 maxBends = [-1] # Call for left subtree of the root if (node.left): findMaxBendsUtil(node.left, 'l', bends, maxBends, 1, Len) # Call for right subtree of the root if (node.right): findMaxBendsUtil(node.right, 'r', bends, maxBends, 1, Len) # Include the root node as well # in the path Length Len[0] += 1 return Len[0]# Driver code if __name__ == '__main__': # Constructed binary tree is # 10 # / \ # 8 2 # / \ / # 3 5 2 # \ # 1 # / # 9 root = newNode(10) root.left = newNode(8) root.right = newNode(2) root.left.left = newNode(3) root.left.right = newNode(5) root.right.left = newNode(2) root.right.left.right = newNode(1) root.right.left.right.left = newNode(9) print(findMaxBends(root) - 1)# This code is contributed by PranchalK |
C#
// C# program to find path length// having maximum number of bendsusing System;public class GFG{ // structure node public class Node { public int key; public Node left; public Node right; }; // Utility function to create a new node static Node newNode(int key) { Node node = new Node(); node.left = null; node.right = null; node.key = key; return node; } /* Recursive function to calculate the pathlength having maximum number of bends.The following are parameters for this function.node -. pointer to the current nodedir -. determines whether the current nodeis left or right child of it's parent nodebends -. number of bends so far in thecurrent path.maxBends -. maximum number of bends in apath from root to leafsoFar -. length of the current path sofar traversedlen -. length of the path having maximumnumber of bends*/ static int maxBends; static int len; static void findMaxBendsUtil(Node node, char dir, int bends, int soFar) { // Base Case if (node == null) return; // Leaf node if (node.left == null && node.right == null) { if (bends > maxBends) { maxBends = bends; len = soFar; } } // Recurring for both left and right child else { if (dir == 'l') { findMaxBendsUtil(node.left, dir, bends, soFar + 1); findMaxBendsUtil(node.right, 'r', bends + 1, soFar + 1); } else { findMaxBendsUtil(node.right, dir, bends, soFar + 1); findMaxBendsUtil(node.left, 'l', bends + 1, soFar + 1); } } } // Helper function to call findMaxBendsUtil() static int findMaxBends(Node node) { if (node == null) return 0; len = 0; maxBends = -1; int bends = 0; // Call for left subtree of the root if (node.left != null) findMaxBendsUtil(node.left, 'l', bends, 1); // Call for right subtree of the root if (node.right != null) findMaxBendsUtil(node.right, 'r', bends, 1); // Include the root node as well in the path length len++; return len; } // Driver code public static void Main(String[] args) { /* Constructed binary tree is 10 / \ 8 2 / \ / 3 5 2 \ 1 / 9 */ Node root = newNode(10); root.left = newNode(8); root.right = newNode(2); root.left.left = newNode(3); root.left.right = newNode(5); root.right.left = newNode(2); root.right.left.right = newNode(1); root.right.left.right.left = newNode(9); Console.Write(findMaxBends(root) - 1); }}// This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript program to find path length // having maximum number of bends class Node { constructor(key) { this.left = null; this.right = null; this.key = key; } } // Utility function to create a new node function newNode(key) { let node = new Node(key); return node; } /* Recursive function to calculate the path length having maximum number of bends. The following are parameters for this function. node -. pointer to the current node dir -. determines whether the current node is left or right child of it's parent node bends -. number of bends so far in the current path. maxBends -. maximum number of bends in a path from root to leaf soFar -. length of the current path so far traversed len -. length of the path having maximum number of bends */ let maxBends; let len; function findMaxBendsUtil(node, dir, bends, soFar) { // Base Case if (node == null) return; // Leaf node if (node.left == null && node.right == null) { if (bends > maxBends) { maxBends = bends; len = soFar; } } // Recurring for both left and right child else { if (dir == 'l') { findMaxBendsUtil(node.left, dir, bends, soFar + 1); findMaxBendsUtil(node.right, 'r', bends + 1, soFar + 1); } else { findMaxBendsUtil(node.right, dir, bends, soFar + 1); findMaxBendsUtil(node.left, 'l', bends + 1, soFar + 1); } } } // Helper function to call findMaxBendsUtil() function findMaxBends(node) { if (node == null) return 0; len = 0; maxBends = -1; let bends = 0; // Call for left subtree of the root if (node.left != null) findMaxBendsUtil(node.left, 'l', bends, 1); // Call for right subtree of the root if (node.right != null) findMaxBendsUtil(node.right, 'r', bends, 1); // Include the root node as well in the path length len++; return len; } /* Constructed binary tree is 10 / \ 8 2 / \ / 3 5 2 \ 1 / 9 */ let root = newNode(10); root.left = newNode(8); root.right = newNode(2); root.left.left = newNode(3); root.left.right = newNode(5); root.right.left = newNode(2); root.right.left.right = newNode(1); root.right.left.right.left = newNode(9); document.write(findMaxBends(root) - 1); </script> |
Output:
4
Time Complexity: O(n) where n is the number of nodes in the binary tree.
Auxiliary Space: O(n)
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