原文:
给定一棵二叉树，求所有叶节点的乘积。 示例:
input :
1
/ \
2 3
/ \ / \
4 5 6 7
\
8
output :
product = 4 * 5 * 8 * 7 = 1120
其思想是以任何方式遍历树，并检查该节点是否是叶节点。如果节点是叶节点，将节点数据乘以一个变量 prod ，用于存储叶节点的乘积。 以下是上述办法的实施情况。
c
// cpp program to find product of
// all leaf nodes of binary tree
#include
using namespace std;
// struct binary tree node
struct node {
int data;
node *left, *right;
};
// return new node
node* newnode(int data)
{
node* temp = new node();
temp->data = data;
temp->left = temp->right = null;
return temp;
}
// utility function which calculates
// product of all leaf nodes
void leafproduct(node* root, int& prod)
{
if (!root)
return;
// product root data to prod if
// root is a leaf node
if (!root->left && !root->right)
prod *= root->data;
// propagate recursively in left
// and right subtree
leafproduct(root->left, prod);
leafproduct(root->right, prod);
}
// driver program
int main()
{
// construct binary tree
node* root = newnode(1);
root->left = newnode(2);
root->left->left = newnode(4);
root->left->right = newnode(5);
root->right = newnode(3);
root->right->right = newnode(7);
root->right->left = newnode(6);
root->right->left->right = newnode(8);
// variable to store product of leaf nodes
int prod = 1;
leafproduct(root, prod);
cout << prod << endl;
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program to find product of
// all leaf nodes of binary tree
class gfg
{
// struct binary tree node
static class node
{
int data;
node left, right;
};
// return new node
static node newnode(int data)
{
node temp = new node();
temp.data = data;
temp.left = temp.right = null;
return temp;
}
// product
static int prod = 1;
// utility function which calculates
// product of all leaf nodes
static void leafproduct(node root )
{
if (root == null)
return;
// product root data to prod if
// root is a leaf node
if (root.left == null && root.right == null)
prod *= root.data;
// propagate recursively in left
// and right subtree
leafproduct(root.left);
leafproduct(root.right);
}
// driver program
public static void main(string args[])
{
// construct binary tree
node root = newnode(1);
root.left = newnode(2);
root.left.left = newnode(4);
root.left.right = newnode(5);
root.right = newnode(3);
root.right.right = newnode(7);
root.right.left = newnode(6);
root.right.left.right = newnode(8);
// variable to store product of leaf nodes
prod = 1;
leafproduct(root);
system.out.println(prod );
}
}
// this code is contributed by arnab kundu
计算机编程语言
# python program to find product of
# all leaf nodes of binary tree
# node class
class node:
# function to initialise the node object
def __init__(self, data):
self.data = data # assign data
self.next =none
# return new node
def newnode( data) :
temp = node(0)
temp.data = data
temp.left = temp.right = none
return temp
# product
prod = 1
# utility function which calculates
# product of all leaf nodes
def leafproduct( root ) :
global prod
if (root == none) :
return
# product root data to prod if
# root is a leaf node
if (root.left == none and root.right == none):
prod *= root.data
# propagate recursively in left
# and right subtree
leafproduct(root.left)
leafproduct(root.right)
# driver program
# construct binary tree
root = newnode(1)
root.left = newnode(2)
root.left.left = newnode(4)
root.left.right = newnode(5)
root.right = newnode(3)
root.right.right = newnode(7)
root.right.left = newnode(6)
root.right.left.right = newnode(8)
# variable to store product of leaf nodes
prod = 1
leafproduct(root)
print(prod )
# this code is contributed by arnab kundu
c
// c# program to find product of
// all leaf nodes of binary tree
using system;
class gfg
{
// struct binary tree node
public class node
{
public int data;
public node left, right;
};
// return new node
static node newnode(int data)
{
node temp = new node();
temp.data = data;
temp.left = temp.right = null;
return temp;
}
// product
static int prod = 1;
// utility function which calculates
// product of all leaf nodes
static void leafproduct(node root )
{
if (root == null)
return;
// product root data to prod if
// root is a leaf node
if (root.left == null && root.right == null)
prod *= root.data;
// propagate recursively in left
// and right subtree
leafproduct(root.left);
leafproduct(root.right);
}
// driver code
public static void main(string []args)
{
// construct binary tree
node root = newnode(1);
root.left = newnode(2);
root.left.left = newnode(4);
root.left.right = newnode(5);
root.right = newnode(3);
root.right.right = newnode(7);
root.right.left = newnode(6);
root.right.left.right = newnode(8);
// variable to store product of leaf nodes
prod = 1;
leafproduct(root);
console.writeline(prod );
}
}
// this code has been contributed by 29ajaykumar
java 描述语言
output:
1120
时间复杂度: o(n)，其中 n 为树中节点总数。
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