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Program to find sum each of the diagonal path elements in a binary tree in Python
Suppose we have a binary tree, we have to find the sum of each of the diagonals in the tree starting from the top to bottom right.
So, if the input is like
then the output will be [27, 18, 3] as the diagonals are [12,15], [8,10],[3]. So the sum values are [27, 18, 3]
To solve this, we will follow these steps −
Define a function traverse() . This will take node, numLeft, output
if node is null, then
return
if numLeft >= size of output , then
insert data of node at the end of output
otherwise,
output[numLeft] := output[numLeft] + data of node
if left of node is not null, then
traverse(left of node, numLeft+1, output)
if right of node is not null, then
traverse(right of node, numLeft, output)
From the main method, do the following −
output := a new list
traverse(root, 0, output)
return output
Let us see the following implementation to get better understanding −
Example
class TreeNode: def __init__(self, data, left = None, right = None): self.data = data self.left = left self.right = right class Solution: def solve(self, root): output = [] def traverse(node, numLeft, output): if not node: return if numLeft >= len(output): output.append(node.data) else: output[numLeft] += node.data if node.left: traverse(node.left, numLeft+1, output) if node.right: traverse(node.right, numLeft, output) traverse(root, 0, output) return output ob = Solution() root = TreeNode(12) root.left = TreeNode(8) root.right = TreeNode(15) root.left.left = TreeNode(3) root.left.right = TreeNode(10) print(ob.solve(root))
Input
root = TreeNode(12) root.left = TreeNode(8) root.right = TreeNode(15) root.left.left = TreeNode(3) root.left.right = TreeNode(10)
Output
[27, 18, 3]
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