Given a binary tree containing n nodes. The problem is to find the sum of all nodes on the longest path from root to leaf node. If two or more paths compete for the longest path, then the path having maximum sum of nodes is being considered.
Input : Binary tree: 4 / \ 2 5 / \ / \ 7 1 2 3 / 6 Output : 13 4 / \ 2 5 / \ / \ 7 1 2 3 / 6 The highlighted nodes (4, 2, 1, 6) above are part of the longest root to leaf path having sum = (4 + 2 + 1 + 6) = 13
Approach: Recursively find the length and sum of nodes of each root to leaf path and accordingly update the maximum sum.
sumOfLongRootToLeafPath(root, sum, len, maxLen, maxSum) if root == NULL if maxLen < len maxLen = len maxSum = sum else if maxLen == len && maxSum is less than sum maxSum = sum return sumOfLongRootToLeafPath(root-left, sum + root-data, len + 1, maxLen, maxSum) sumOfLongRootToLeafPath(root-right, sum + root-data, len + 1, maxLen, maxSum) sumOfLongRootToLeafPathUtil(root) if (root == NULL) return 0 Declare maxSum = Minimum Integer Declare maxLen = 0 sumOfLongRootToLeafPath(root, 0, 0, maxLen, maxSum) return maxSum
Sum = 13
Time Complexity: O(n)
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