原文:
给定一个二叉树。任务是查找并打印树中所有内部节点(非叶节点)的乘积和总和。
在上面的树中，只有两个节点 1 和 2 是非叶节点。 因此，非叶节点的积= 1 * 2 = 2。 非叶节点之和= 1 2 =3。 例:
input :
1
/ \
2 3
/ \ / \
4 5 6 7
\
8
output : product = 36, sum = 12
non-leaf nodes are: 1, 2, 3, 6
方法:想法是以任何方式遍历树，并检查当前节点是否是非叶节点。取两个变量积和和分别存储非叶节点的积和。如果当前节点是非叶节点，则将该节点的数据乘以用于存储非叶节点乘积的变量乘积，并将该节点的数据加到用于存储非叶节点总和的变量和上。 以下是上述思路的实现:
c
// cpp program to find product and sum of
// non-leaf 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;
struct node* right;
};
/* 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);
}
// computes the product of non-leaf
// nodes in a tree
void findproductsum(struct node* root, int& prod, int& sum)
{
// base cases
if (root == null || (root->left == null
&& root->right == null))
return;
// if current node is non-leaf,
// calculate product and sum
if (root->left != null || root->right != null)
{
prod *= root->data;
sum = root->data;
}
// if root is not null and its one of its
// child is also not null
findproductsum(root->left, prod, sum);
findproductsum(root->right, prod, sum);
}
// driver code
int main()
{
// binary tree
struct node* root = newnode(1);
root->left = newnode(2);
root->right = newnode(3);
root->left->left = newnode(4);
root->left->right = newnode(5);
int prod = 1;
int sum = 0;
findproductsum(root, prod, sum);
cout <<"product = "<// java program to find product and sum of
// non-leaf nodes in a binary tree
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;
node right;
};
/* 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 = node.right = null;
return (node);
}
//int class
static class int
{
int a;
}
// computes the product of non-leaf
// nodes in a tree
static void findproductsum(node root, int prod, int sum)
{
// base cases
if (root == null || (root.left == null
&& root.right == null))
return;
// if current node is non-leaf,
// calculate product and sum
if (root.left != null || root.right != null)
{
prod.a *= root.data;
sum.a = root.data;
}
// if root is not null and its one of its
// child is also not null
findproductsum(root.left, prod, sum);
findproductsum(root.right, prod, sum);
}
// driver code
public static void main(string args[])
{
// binary tree
node root = newnode(1);
root.left = newnode(2);
root.right = newnode(3);
root.left.left = newnode(4);
root.left.right = newnode(5);
int prod = new int();prod.a = 1;
int sum = new int(); sum.a = 0;
findproductsum(root, prod, sum);
system.out.print("product = " prod.a " , sum = " sum.a);
}
}
// this code is contributed by arnab kundu
python 3
# python3 program to find product and sum
# of non-leaf nodes in a binary tree
# helper function that allocates a new
# node with the given data and none
# left and right poers.
class newnode:
# construct to create a new node
def __init__(self, key):
self.data = key
self.left = none
self.right = none
# computes the product of non-leaf
# nodes in a tree
class new:
def findproductsum(sf,root) :
# base cases
if (root == none or (root.left == none and
root.right == none)) :
return
# if current node is non-leaf,
# calculate product and sum
if (root.left != none or
root.right != none) :
sf.prod *= root.data
sf.sum = root.data
# if root is not none and its one
# of its child is also not none
sf.findproductsum(root.left)
sf.findproductsum(root.right)
def main(sf):
root = newnode(1)
root.left = newnode(2)
root.right = newnode(3)
root.left.left = newnode(4)
root.left.right = newnode(5)
sf.prod = 1
sf.sum = 0
sf.findproductsum(root)
print("product =", sf.prod,
", sum =", sf.sum)
# driver code
if __name__ == '__main__':
x = new()
x.main()
# this code is contributed by
# shubham singh(shubhamsingh10)
c
// c# program to find product and sum of
// non-leaf 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;
public node right;
};
/* 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 = node.right = null;
return (node);
}
// int class
public class int
{
public int a;
}
// computes the product of non-leaf
// nodes in a tree
static void findproductsum(node root, int prod, int sum)
{
// base cases
if (root == null || (root.left == null
&& root.right == null))
return;
// if current node is non-leaf,
// calculate product and sum
if (root.left != null || root.right != null)
{
prod.a *= root.data;
sum.a = root.data;
}
// if root is not null and its one of its
// child is also not null
findproductsum(root.left, prod, sum);
findproductsum(root.right, prod, sum);
}
// driver code
public static void main(string []args)
{
// binary tree
node root = newnode(1);
root.left = newnode(2);
root.right = newnode(3);
root.left.left = newnode(4);
root.left.right = newnode(5);
int prod = new int();prod.a = 1;
int sum = new int(); sum.a = 0;
findproductsum(root, prod, sum);
console.write("product = " prod.a " , sum = " sum.a);
}
}
// this code is contributed by 29ajaykumar
java 描述语言
output:
product = 2 , sum = 3
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