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Suppose we have a binary tree; we have to find the number of nodes that are an only child. As we know a node x is called an only child node when its parent has exactly one child that is x.

So, if the input is like

then the output will be 2 as 8 and 6 are the only children.

To solve this, we will follow these steps:

- if root is null, then
- return 0

- d := a double ended queue
- insert root at the end of d
- count := 0
- while d is not empty, do
- current := left element of d and delete left element
- if left of current is not null, then
- insert left of current into d
- if right of current is null, then
- count := count + 1

- if right of current is not null, then
- insert right of current into d
- if left of current is null, then
- count := count + 1

- return count

Let us see the following implementation to get better understanding:

from collections import deque 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): if not root: return 0 d = deque() d.append(root) count = 0 while d: current = d.popleft() if current.left: d.append(current.left) if not current.right: count += 1 if current.right: d.append(current.right) if not current.left: count += 1 return count ob = Solution() root = TreeNode(9) root.left = TreeNode(7) root.right = TreeNode(10) root.left.right = TreeNode(8) root.right.right = TreeNode(6) print(ob.solve(root))

root = TreeNode(9)root.left = TreeNode(7)

root.right = TreeNode(10)

root.left.right = TreeNode(8)

root.right.right = TreeNode(6)

2

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