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Suppose we have a binary tree, we have to find a list of two numbers where the first number is the count of leaves in the tree and the second number is the count of non-leaf nodes.

So, if the input is like

then the output will be (3, 2), as there are 3 leaves and 2 non-leaf nodes.

To solve this, we will follow these steps −

- if n is null, then
- return (0, 0)

- if left of n is null and right of n is null, then
- return (1, 0)

- left := solve(left of n)
- right := solve(right of n)
- return (left[0] + right[0], 1 + left[1] + right[1])

Let us see the following implementation to get better understanding −

class TreeNode: def __init__(self, data, left = None, right = None): self.val = data self.left = left self.right = right class Solution: def solve(self, n): if not n: return 0, 0 if not n.left and not n.right: return 1, 0 left, right = self.solve(n.left), self.solve(n.right) return left[0] + right[0], 1 + left[1] + right[1] ob = Solution() root = TreeNode(6) root.left = TreeNode(2) root.right = TreeNode(6) root.right.left = TreeNode(10) root.right.right = TreeNode(2) print(ob.solve(root))

root = TreeNode(6) root.left = TreeNode(2) root.right = TreeNode(6) root.right.left = TreeNode(10) root.right.right = TreeNode(2)

(3, 2)

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