计算二叉树中的一半节点（迭代和递归）

给定一棵二叉树，如何在不使用递归的情况下计算所有半节点（只有一个子节点）？注意叶子不应该被触摸，因为它们的两个孩子都为 NULL。

Input : Root of below tree

Output : 3
Nodes 7, 5 and 9 are half nodes as one of 
their child is Null. So count of half nodes
in the above tree is 3

迭代
这个想法是使用水平顺序遍历来有效地解决这个问题。

1) Create an empty Queue Node and push root node to Queue.
2) Do following while nodeQeue is not empty.
   a) Pop an item from Queue and process it.
      a.1) If it is half node then increment count++.
   b) Push left child of popped item to Queue, if available.
   c) Push right child of popped item to Queue, if available.

下面是这个想法的实现。

C++

// C++ program to count half nodes in a Binary Tree
#include 
using namespace std;
 
// A binary tree Node has data, pointer to left
// child and a pointer to right child
struct Node
{
    int data;
    struct Node* left, *right;
};
 
// Function to get the count of half Nodes in
// a binary tree
unsigned int gethalfCount(struct Node* node)
{
    // If tree is empty
    if (!node)
        return 0;
 
    int count = 0; // Initialize count of half nodes
 
    // Do level order traversal starting from root
    queue q;
    q.push(node);
    while (!q.empty())
    {
        struct Node *temp = q.front();
        q.pop();
 
        if (!temp->left && temp->right ||
            temp->left && !temp->right)
            count++;
 
        if (temp->left != NULL)
            q.push(temp->left);
        if (temp->right != NULL)
            q.push(temp->right);
    }
    return count;
}
 
/* Helper function that allocates a new
   Node with the given data and NULL left
   and right pointers. */
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}
 
// Driver program
int main(void)
{
    /* 2
     / \
    7     5
    \     \
     6     9
    / \ /
    1 11 4
    Let us create Binary Tree shown in
    above example */
 
    struct Node *root = newNode(2);
    root->left     = newNode(7);
    root->right     = newNode(5);
    root->left->right = newNode(6);
    root->left->right->left = newNode(1);
    root->left->right->right = newNode(11);
    root->right->right = newNode(9);
    root->right->right->left = newNode(4);
 
    cout << gethalfCount(root);
 
    return 0;
}

Java

// Java program to count half nodes in a Binary Tree
// using Iterative approach
import java.util.Queue;
import java.util.LinkedList;
 
// Class to represent Tree node
class Node
{
    int data;
    Node left, right;
 
    public Node(int item)
    {
        data = item;
        left = null;
        right = null;
    }
}
 
// Class to count half nodes of Tree
class BinaryTree
{
 
    Node root;
 
    /* Function to get the count of half Nodes in
    a binary tree*/
    int gethalfCount()
    {
        // If tree is empty
        if (root==null)
            return 0;
 
        // Do level order traversal starting from root
        Queue queue = new LinkedList();
        queue.add(root);
 
        int count=0; // Initialize count of half nodes
        while (!queue.isEmpty())
        {
 
            Node temp = queue.poll();
            if (temp.left!=null && temp.right==null ||
                temp.left==null && temp.right!=null)
                count++;
 
            // Enqueue left child
            if (temp.left != null)
                queue.add(temp.left);
 
            // Enqueue right child
            if (temp.right != null)
                queue.add(temp.right);
        }
        return count;
    }
 
    public static void main(String args[])
    {
        /* 2
          / \
        7     5
        \     \
        6     9
        / \ /
        1 11 4
        Let us create Binary Tree shown in
        above example */
        BinaryTree tree_level = new BinaryTree();
        tree_level.root = new Node(2);
        tree_level.root.left = new Node(7);
        tree_level.root.right = new Node(5);
        tree_level.root.left.right = new Node(6);
        tree_level.root.left.right.left = new Node(1);
        tree_level.root.left.right.right = new Node(11);
        tree_level.root.right.right = new Node(9);
        tree_level.root.right.right.left = new Node(4);
 
        System.out.println(tree_level.gethalfCount());
 
    }
}

Python3

# Python program to count
# half nodes in a Binary Tree
# using iterative approach
 
# A node structure
class Node:
 
    # A utility function to create a new node
    def __init__(self ,key):
        self.data = key
        self.left = None
        self.right = None
 
# Iterative Method to count half nodes of binary tree
def gethalfCount(root):
 
    # Base Case
    if root is None:
        return 0
 
    # Create an empty queue for level order traversal
    queue = []
 
    # Enqueue Root and initialize count
    queue.append(root)
 
    count = 0 #initialize count for half nodes
    while(len(queue) > 0):
 
        node = queue.pop(0)
 
        # if it is half node then increment count
        if node.left is not None and node.right is None or node.left is None and node.right is not None:
            count = count+1
 
        #Enqueue left child
        if node.left is not None:
            queue.append(node.left)
 
        # Enqueue right child
        if node.right is not None:
            queue.append(node.right)
 
    return count
 
#Driver Program to test above function
 
root = Node(2)
root.left = Node(7)
root.right = Node(5)
root.left.right = Node(6)
root.left.right.left = Node(1)
root.left.right.right = Node(11)
root.right.right = Node(9)
root.right.right.left = Node(4)
 
 
print "%d" %(gethalfCount(root))

C#

// C# program to count half nodes in a Binary Tree
// using Iterative approach
using System;
using System.Collections.Generic;
 
// Class to represent Tree node
public class Node
{
    public int data;
    public Node left, right;
 
    public Node(int item)
    {
        data = item;
        left = null;
        right = null;
    }
}
 
// Class to count half nodes of Tree
public class BinaryTree
{
 
    Node root;
 
    /* Function to get the count of half Nodes in
    a binary tree*/
    int gethalfCount()
    {
        // If tree is empty
        if (root == null)
            return 0;
 
        // Do level order traversal starting from root
        Queue queue = new Queue();
        queue.Enqueue(root);
 
        int count = 0; // Initialize count of half nodes
        while (queue.Count != 0)
        {
 
            Node temp = queue.Dequeue();
            if (temp.left != null && temp.right == null ||
                temp.left == null && temp.right != null)
                count++;
 
            // Enqueue left child
            if (temp.left != null)
                queue.Enqueue(temp.left);
 
            // Enqueue right child
            if (temp.right != null)
                queue.Enqueue(temp.right);
        }
        return count;
    }
 
    // Driver code
    public static void Main()
    {
        /* 2
        / \
        7 5
        \ \
        6 9
        / \ /
        1 11 4
        Let us create Binary Tree shown in
        above example */
        BinaryTree tree_level = new BinaryTree();
        tree_level.root = new Node(2);
        tree_level.root.left = new Node(7);
        tree_level.root.right = new Node(5);
        tree_level.root.left.right = new Node(6);
        tree_level.root.left.right.left = new Node(1);
        tree_level.root.left.right.right = new Node(11);
        tree_level.root.right.right = new Node(9);
        tree_level.root.right.right.left = new Node(4);
 
        Console.WriteLine(tree_level.gethalfCount());
    }
}
 
// This code contributed by Rajput-Ji

Javascript

 
// Javascript program to count half nodes in a Binary Tree
// using Iterative approach
 
// Class to represent Tree node
class Node
{
    constructor(item)
    {
        this.data = item;
        this.left = null;
        this.right = null;
    }
}
 
// Class to count half nodes of Tree
var root;
 
/* Function to get the count of half Nodes in
a binary tree*/
function gethalfCount()
{
 
    // If tree is empty
    if (root == null)
        return 0;
         
    // Do level order traversal starting from root
    var queue = [];
    queue.push(root);
    var count = 0; // Initialize count of half nodes
    while (queue.length != 0)
    {
        var temp = queue.shift();
        if (temp.left != null && temp.right == null ||
            temp.left == null && temp.right != null)
            count++;
             
        // push left child
        if (temp.left != null)
            queue.push(temp.left);
             
        // push right child
        if (temp.right != null)
            queue.push(temp.right);
    }
    return count;
}
 
// Driver code
/* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree shown in
above example */
root = new Node(2);
root.left = new Node(7);
root.right = new Node(5);
root.left.right = new Node(6);
root.left.right.left = new Node(1);
root.left.right.right = new Node(11);
root.right.right = new Node(9);
root.right.right.left = new Node(4);
document.write(gethalfCount());
 
// This code is contributed by famously.

输出：

3

时间复杂度： O(n)
辅助空间： O(n)
其中，n 是给定二叉树中的节点数

递归的
这个想法是按后序遍历树。如果当前节点是一半，我们将结果加 1 并添加左右子树的返回值。

C++

// C++ program to count half nodes in a Binary Tree
#include 
using namespace std;
 
// A binary tree Node has data, pointer to left
// child and a pointer to right child
struct Node
{
    int data;
    struct Node* left, *right;
};
 
// Function to get the count of half Nodes in
// a binary tree
unsigned int gethalfCount(struct Node* root)
{
    if (root == NULL)
       return 0;
 
    int res = 0;
    if  ((root->left == NULL && root->right != NULL) ||
         (root->left != NULL && root->right == NULL))
       res++;
 
    res += (gethalfCount(root->left) + gethalfCount(root->right));
    return res;
}
 
/* Helper function that allocates a new
   Node with the given data and NULL left
   and right pointers. */
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}
 
// Driver program
int main(void)
{
    /* 2
     / \
    7     5
    \     \
     6     9
    / \ /
    1 11 4
    Let us create Binary Tree shown in
    above example */
 
    struct Node *root = newNode(2);
    root->left     = newNode(7);
    root->right     = newNode(5);
    root->left->right = newNode(6);
    root->left->right->left = newNode(1);
    root->left->right->right = newNode(11);
    root->right->right = newNode(9);
    root->right->right->left = newNode(4);
 
    cout << gethalfCount(root);
 
    return 0;
}

Java

// Java program to count half nodes in a Binary Tree
import java.util.*;
class GfG {
 
// A binary tree Node has data, pointer to left
// child and a pointer to right child
static class Node
{
    int data;
    Node left, right;
}
 
// Function to get the count of half Nodes in
// a binary tree
static int gethalfCount(Node root)
{
    if (root == null)
    return 0;
 
    int res = 0;
    if ((root.left == null && root.right != null) ||
        (root.left != null && root.right == null))
    res++;
 
    res += (gethalfCount(root.left)
            + gethalfCount(root.right));
    return res;
}
 
/* Helper function that allocates a new
Node with the given data and NULL left
and right pointers. */
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = null;
    node.right = null;
    return (node);
}
 
// Driver program
public static void main(String[] args)
{
    /* 2
    / \
    7 5
    \ \
    6 9
    / \ /
    1 11 4
    Let us create Binary Tree shown in
    above example */
 
    Node root = newNode(2);
    root.left = newNode(7);
    root.right = newNode(5);
    root.left.right = newNode(6);
    root.left.right.left = newNode(1);
    root.left.right.right = newNode(11);
    root.right.right = newNode(9);
    root.right.right.left = newNode(4);
 
    System.out.println(gethalfCount(root));
 
}
}

Python3

# Python program to count half nodes in a binary tree
 
# A node structure
class newNode:
    def __init__(self, data):
        self.data = data
        self.left = self.right = None
 
# Function to get the count of half Nodes in a binary tree
def gethalfCount(root):
    if root == None:
        return 0
    res = 0
    if(root.left == None and root.right != None) or \
        (root.left != None and root.right == None):
        res += 1
    res += (gethalfCount(root.left) + \
            gethalfCount(root.right))
    return res
 
# Driver program
''' 2
    / \
    7     5
    \     \
    6     9
    / \ /
    1 11 4
    Let us create Binary Tree shown in
    above example '''
 
root = newNode(2)
root.left = newNode(7)
root.right = newNode(5)
root.left.right = newNode(6)
root.left.right.left = newNode(1)
root.left.right.right = newNode(11)
root.right.right = newNode(9)
root.right.right.left = newNode(4)
print(gethalfCount(root))
 
# This code is contributed by simranjenny84

C＃

// C# program to count half nodes in a Binary Tree
using System;
 
class GfG
{
 
    // A binary tree Node has data, pointer to left
    // child and a pointer to right child
    public class Node
    {
        public int data;
        public Node left, right;
    }
 
    // Function to get the count of half Nodes in
    // a binary tree
    static int gethalfCount(Node root)
    {
        if (root == null)
        return 0;
 
        int res = 0;
        if ((root.left == null && root.right != null) ||
            (root.left != null && root.right == null))
        res++;
 
        res += (gethalfCount(root.left)
                + gethalfCount(root.right));
        return res;
    }
 
    /* Helper function that allocates a new
    Node with the given data and NULL left
    and right pointers. */
    static Node newNode(int data)
    {
        Node node = new Node();
        node.data = data;
        node.left = null;
        node.right = null;
        return (node);
    }
 
    // Driver code
    public static void Main()
    {
        /* 2
        / \
        7 5
        \ \
        6 9
        / \ /
        1 11 4
        Let us create Binary Tree shown in
        above example */
 
        Node root = newNode(2);
        root.left = newNode(7);
        root.right = newNode(5);
        root.left.right = newNode(6);
        root.left.right.left = newNode(1);
        root.left.right.right = newNode(11);
        root.right.right = newNode(9);
        root.right.right.left = newNode(4);
 
        Console.WriteLine(gethalfCount(root));
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

 
// Javascript program to count half
// nodes in a Binary Tree
class Node
{
    constructor(data)
    {
        this.left = null;
        this.right = null;
        this.data = data;
    }
}
 
// Function to get the count of half
// Nodes in a binary tree
function gethalfCount(root)
{
    if (root == null)
        return 0;
 
    let res = 0;
     
    if ((root.left == null && root.right != null) ||
        (root.left != null && root.right == null))
        res++;
 
    res += (gethalfCount(root.left) +
            gethalfCount(root.right));
    return res;
}
 
/* Helper function that allocates a new
Node with the given data and NULL left
and right pointers. */
function newNode(data)
{
    let node = new Node(data);
    return (node);
}
 
// Driver code
/*          2
        / \
        7  5
        \   \
         6    9
        / \  /
       1  11 4
Let us create Binary Tree shown in
above example */
let root = newNode(2);
root.left = newNode(7);
root.right = newNode(5);
root.left.right = newNode(6);
root.left.right.left = newNode(1);
root.left.right.right = newNode(11);
root.right.right = newNode(9);
root.right.right.left = newNode(4);
 
document.write(gethalfCount(root));
 
// This code is contributed by divyeshrabadiya07
 

输出 ：

3

时间复杂度： O(n)
辅助空间： O(n)
其中，n 是给定二叉树中的节点数

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