给定一棵二叉树，任务是打印具有一个孩子的所有节点。如果不存在这样的节点，则打印“ -1”。
例子：
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.
方法：想法是按顺序遍历树，并在遍历的每个步骤中检查该节点是否正好有一个孩子。然后将该节点追加到结果数组中以跟踪此类节点。遍历后，只需打印此结果数组的每个元素。
下面是上述方法的实现：
C++
// C++ implementation to print
// the nodes having a single child
#include
using namespace std;
// Class of the Binary Tree node
struct Node
{
int data;
Node *left, *right;
Node(int x)
{
data = x;
left = right = NULL;
}
};
vector lst;
// Function to find the nodes
// having single child
void printNodesOneChild(Node* root)
{
// Base case
if (root == NULL)
return;
// Condition to check if the
// node is having only one child
if (root->left != NULL &&
root->right == NULL ||
root->left == NULL &&
root->right != NULL)
{
lst.push_back(root->data);
}
// Traversing the left child
printNodesOneChild(root -> left);
// Traversing the right child
printNodesOneChild(root -> right);
}
//Driver code
int main()
{
// Constructing the binary tree
Node *root = new Node(2);
root -> left = new Node(3);
root -> right = new Node(5);
root -> left -> left = new Node(7);
root -> right -> left = new Node(8);
root -> right -> right = new Node(6);
// Function calling
printNodesOneChild(root);
// Condition to check if there is
// no such node having single child
if (lst.size() == 0)
printf("-1");
else
{
for(int value : lst)
{
cout << (value) << endl;
}
}
}
// This code is contributed by Mohit Kumar 29 Java
// Java implementation to print
// the nodes having a single child
import java.util.Vector;
// Class of the Binary Tree node
class Node
{
int data;
Node left, right;
Node(int data)
{
this.data = data;
}
}
class GFG{
// List to keep track of nodes
// having single child
static Vector list = new Vector<>();
// Function to find the nodes
// having single child
static void printNodesOneChild(Node root)
{
// Base case
if (root == null)
return;
// Condition to check if the node
// is having only one child
if (root.left != null && root.right == null ||
root.left == null && root.right != null)
{
list.add(root.data);
}
// Traversing the left child
printNodesOneChild(root.left);
// Traversing the right child
printNodesOneChild(root.right);
}
// Driver code
public static void main(String[] args)
{
// Constructing the binary tree
Node root = new Node(2);
root.left = new Node(3);
root.right = new Node(5);
root.left.left = new Node(7);
root.right.left = new Node(8);
root.right.right = new Node(6);
// Function calling
printNodesOneChild(root);
// Condition to check if there is
// no such node having single child
if (list.size() == 0)
System.out.println(-1);
else
{
for(int value : list)
{
System.out.println(value);
}
}
}
}
// This code is contributed by sumit_9 Python3
# Python3 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()C#
// C# implementation to print
// the nodes having a single child
using System;
using System.Collections;
class GFG{
// Structure of binary tree
public class Node
{
public Node left;
public Node right;
public int data;
};
// Function to create a new node
static Node newNode(int key)
{
Node node = new Node();
node.left = node.right = null;
node.data = key;
return node;
}
// List to keep track of nodes
// having single child
static ArrayList list = new ArrayList();
// Function to find the nodes
// having single child
static void printNodesOneChild(Node root)
{
// Base case
if (root == null)
return;
// Condition to check if the node
// is having only one child
if (root.left != null && root.right == null ||
root.left == null && root.right != null)
{
list.Add(root.data);
}
// Traversing the left child
printNodesOneChild(root.left);
// Traversing the right child
printNodesOneChild(root.right);
}
// Driver code
static public void Main()
{
// Constructing the binary tree
Node root = newNode(2);
root.left = newNode(3);
root.right = newNode(5);
root.left.left = newNode(7);
root.right.left = newNode(8);
root.right.right = newNode(6);
// Function calling
printNodesOneChild(root);
// Condition to check if there is
// no such node having single child
if (list.Count == 0)
Console.WriteLine(-1);
else
{
foreach(int value in list)
{
Console.WriteLine(value);
}
}
}
}
// This code is contributed by offbeat输出
3