The time complexity of the above code is O(2^N) The main idea to solve this problem is to use Backtracking. The recursion tree is where we have skipped the 3rd and 6th characters.
My solution would be to traverse the tree, and for every node, find the maximum value node in its left and right subtree. If the difference between the node and its descendants is more than the maximum difference found so far, update it. The time complexity of this solution is O(n^2), where n is the total number of nodes in the binary tree.
If the root is NULL, return 0(Base Case) Call the recursive function to find the max sum for the left and the right subtree In a variable store the maximum of (root->data, maximum of (leftSum, rightSum) + root->data) In another variable store the maximum of previous step and root->data + leftSum + rightSum Return the maximum of the previous step The time complexity of the above code is O(N)
Time Complexity: O(N) Using DFS: Always multiply temp by 10 till temp * 10 is greater than n Increment temp by 1 when the last digit of temp is not equal to 9