# Print the nodes having exactly one child in a Binary tree

Given a binary tree, the task is to print all the nodes having exactly one child. Print “-1” if no such node exists.

Examples:

Input:
2
/ \
3   5
/   / \
7   8   6
Output: 3
Explanation:
There is only one node having
single child that is 3.

Input:
9
/ \
7   8
/ \
4   3
Output: -1
Explanation:
There is no node having exactly one
child in the binary tree.

Approach: The idea is to traverse the tree in the inorder traversal and at each step of the traversal check that if the node is having exactly one child. Then append that node into a result array to keep track of such nodes. After the traversal, simply print each element of this result array.

Below is the implementation of the above approach:

## Python

 # Python implementation to print # the nodes having a single child    # Class of the Binary Tree node class node:            # Constructor to construct      # the node of the Binary tree     def __init__(self, val):         self.val = val         self.left = None         self.right = None            # List to keep track of  # nodes having single child single_child_nodes = []    # Function to find the nodes  # having single child  def printNodesOneChild(root):            # Base Case     if not root:         return            # Condition to check if the node     # is having only one child     if not root.left and root.right:         single_child_nodes.append(root)     elif root.left and not root.right:         single_child_nodes.append(root)                # Traversing the left child     printNodesOneChild(root.left)            # Traversing the right child     printNodesOneChild(root.right)     return    # Driver Code if __name__ == "__main__":            # Condtruction of Binary Tree     root = node(2)     root.left = node(3)     root.right = node(5)     root.left.left = node(7)     root.right.left = node(8)     root.right.right = node(6)            # Function Call     printNodesOneChild(root)            # Condition to check if there is      # no such node having single child     if not len(single_child_nodes):         print(-1)     else:         for i in single_child_nodes:              print(i.val, end = " ")         print()

Output:

3

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