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Maximum Average Subtree in Python
Suppose we have the root of a binary tree; we have to find the maximum average value of any subtree of that tree. So if the tree is like −
The output will be 6, this is because, for node 5, it will be (5 + 6 + 1)/ 3 = 4, then for node 6 it will be 6 / 1 = 6, and for node 1, it will be 1 / 1 = 1, so the max is 6.
To solve this, we will follow these steps −
res := 0
Define a method called solve(), this will take root
if root is not set, then return a pair [0, 0]
left := solve(left of root) right := solve(right of root)
c := left[0] + right[0] + 1
s := left[1] + right[1] + value of root
ans := max of and ans s/c
return a pair[c, s]
From the main method, set ans := 0, call solve(root) and return ans
Example(Python)
Let us see the following implementation to get better understanding −
class TreeNode: def __init__(self, data, left = None, right = None): self.data = data self.left = left self.right = right def insert(temp,data): que = [] que.append(temp) while (len(que)): temp = que[0] que.pop(0) if (not temp.left): if data is not None: temp.left = TreeNode(data) else: temp.left = TreeNode(0) break else: que.append(temp.left) if (not temp.right): if data is not None: temp.right = TreeNode(data) else: temp.right = TreeNode(0) break else: que.append(temp.right) def make_tree(elements): Tree = TreeNode(elements[0]) for element in elements[1:]: insert(Tree, element) return Tree class Solution(object): def helper(self, node): if not node: return 0, 0 left_sum, left_count = self.helper(node.left) right_sum, right_count = self.helper(node.right) self.ans = max(self.ans, (left_sum + right_sum + node.data) / (left_count + right_count + 1)) return left_sum + right_sum + node.data, left_count + right_count + 1 def maximumAverageSubtree(self, root): self.ans = 0 self.helper(root) return self.ans ob = Solution() root = make_tree([5,6,1]) print(ob.maximumAverageSubtree(root))
Input
[5,6,1]
Output
6.0
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