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Maximum Consecutive Increasing Path Length in Binary Tree in C++
Suppose we have a binary tree; we have to calculate the length of the longest path which consists of nodes with consecutive values in increasing order. Every node will be treated as a path of length 1.
So, if the input is like
then the output will be 3 as (11, 12, 13) is maximum consecutive path.
To solve this, we will follow these steps −
- Define a function solve(), this will take root, prev_data, prev_length,
- if not root is non-zero, then −
- return prev_length
- cur_data := val of root
- if cur_data is same as prev_data + 1, then −
- return maximum of solve(left of root, cur_data, prev_length+1) and solve(right of root, cur_data, prev_length+1)
- newPathLen := maximum of solve(left of root, cur_data, 1) and solve(right of root, cur_data, 1)
- return maximum of prev_length and newPathLen
- From the main method do the following −
- if root is same as NULL, then −
- return 0
- return solve(root, val of root-1, 0)
Example (C++)
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h> using namespace std; class TreeNode { public: int val; TreeNode *left, *right; TreeNode(int data) { val = data; left = NULL; right = NULL; } }; int solve(TreeNode *root, int prev_data, int prev_length){ if (!root) return prev_length; int cur_data = root->val; if (cur_data == prev_data+1){ return max(solve(root->left, cur_data, prev_length+1), solve(root->right, cur_data, prev_length+1)); } int newPathLen = max(solve(root->left, cur_data, 1), solve(root->right, cur_data, 1)); return max(prev_length, newPathLen); } int maxLen(TreeNode *root){ if (root == NULL) return 0; return solve(root, root->val-1, 0); } int main(){ TreeNode *root = new TreeNode(10); root->left = new TreeNode(11); root->right = new TreeNode(9); root->left->left = new TreeNode(13); root->left->right = new TreeNode(12); root->right->left = new TreeNode(13); root->right->right = new TreeNode(8); cout << maxLen(root); return 0; }
Input
TreeNode *root = new TreeNode(10); root->left = new TreeNode(11); root->right = new TreeNode(9); root->left->left = new TreeNode(13); root->left->right = new TreeNode(12); root->right->left = new TreeNode(13); root->right->right = new TreeNode(8);
Output
3
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