给定两个二叉树和两个整数密钥K1、K2 ，其中K1、K2可以出现在同一棵树或不同的树中。设F1, F2是表示从根到K1和K2 (不包括 K1 和 K2)的顶点序列的向量，任务是找到这两个向量的点积。
注意：树中没有重复项，并且两棵树彼此不同。如果键存在于不同的深度，则只考虑节点直到相同的深度级别。
例子：
Input: K1 = 4, K2 = 5
Output: 5
Explanation:
Clearly from the 2 trees, both the keys are present in the first tree. The sequence of the vertices from the root to the K1 or F1 vector is, F1 = (1, 2) and F2 = (1, 2).
Now the dot product i.e. F1.F2 = 1×1 + 2×2 = 5.
Input: K1 = 5, K2 = 7
Output: 6
Explanation:
F1 = (1, 2) and F2 = (6)
Since only need to consider the nodes till the same depth level, the dot product is F1.F2 = 1×6 = 6.
方法：想法是找到存在给定键值的树，然后计算祖先的点积。请按照以下步骤解决问题：
- 初始化两个不同的辅助向量。
- 将两个键的祖先存储在向量中。
- 遍历向量，直到到达任意一个向量的末尾，继续计算向量对应元素的点积。
- 打印最终的点积作为答案。
C++
// C++ program for the above approach
#include
using namespace std;
// Structure of the tree node
struct node {
int data;
struct node* left;
struct node* right;
};
// Utility function to create a new node
struct node* newNode(int data)
{
struct node* node
= (struct node*)malloc(
sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
return (node);
}
// Function to store the ancestors
// of the given key in the vector
bool printAncestors(struct node* root,
int K, vector& v)
{
// Base case
if (root == NULL)
return false;
if (root->data == K)
return true;
// If target is present in either left
// or right subtree of this node,
// then print this node
if (printAncestors(root->left, K, v)
|| printAncestors(root->right, K, v)) {
v.push_back(root->data);
return true;
}
return false;
}
// Function to store the dot product of the vectors
int dotProductOfVectors(vector& v1, vector& v2)
{
// Traverse the vectors from the end because the
// ancestors starts from the root of the
// tree and root of the tree is present at the end
int size1 = v1.size();
int size2 = v2.size();
int i = size1 - 1;
int j = size2 - 1;
int answer = 0;
// Traverse the vectors side by side and storing answer
while (i >= 0 && j >= 0) {
answer = answer + (v1[i] * v2[j]);
i--;
j--;
}
// Return dot product
return answer;
}
// Utility function to calculate the dot product of
// the ancestors of the keys
int dotProductOfAncestorsUtil(struct node* root1,
struct node* root2, int K1,
int K2)
{
// To store the ancestors of each key
vector F1, F2;
// If both keys are present in the first tree
if (printAncestors(root1, K1, F1)
&& printAncestors(root1, K2, F2)) {
return dotProductOfVectors(F1, F2);
}
// If both keys are present in the second tree
else if (printAncestors(root2, K1, F1)
&& printAncestors(root2, K2, F2)) {
return dotProductOfVectors(F1, F2);
}
// If first key exists in first tree and
// the second key exists in second tree
else if (printAncestors(root1, K1, F1)
&& printAncestors(root2, K2, F2)) {
return dotProductOfVectors(F1, F2);
}
// Otherwise, first key exists in second tree and
// the second key exists in first tree
else {
if (printAncestors(root2, K1, F1)
&& printAncestors(root1, K2, F2)) {
return dotProductOfVectors(F1, F2);
}
}
// If either of the nodes doesn't exist
return 0;
}
void dotProductOfAncestors(struct node* root1,
struct node* root2, int K1,
int K2)
{
// To store dot product of two vwctors
int dotProduct
= dotProductOfAncestorsUtil(root1, root2, K1, K2);
// Print dot product as the answer
cout << dotProduct;
}
// Driver Code
int main()
{
/* Construct the following binary trees
1 6
/ \ / \
2 3 7 8
/ \ / \
4 5 9 10
*/
// Given Tree 1
struct node* root1 = newNode(1);
root1->left = newNode(2);
root1->right = newNode(3);
root1->left->left = newNode(4);
root1->left->right = newNode(5);
// Given Tree 2
struct node* root2 = newNode(6);
root2->left = newNode(7);
root2->right = newNode(8);
root2->left->left = newNode(9);
root2->right->right = newNode(10);
// Given keys
int K1 = 4, K2 = 5;
// Function Call
dotProductOfAncestors(root1, root2, K1, K2);
} Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG
{
// Structure of the tree node
static class node
{
int data;
node left, right;
}
// Utility function to create a new node
static node newNode(int data)
{
node node = new node();
node.data = data;
node.left = null;
node.right = null;
return (node);
}
// Function to store the ancestors
// of the given key in the vector
static boolean printAncestors(node root, int K,
Vector v)
{
// Base case
if (root == null)
return false;
if (root.data == K)
return true;
// If target is present in either left
// or right subtree of this node,
// then print this node
if (printAncestors(root.left, K, v)
|| printAncestors(root.right, K, v))
{
v.add(root.data);
return true;
}
return false;
}
// Function to store the dot product of the vectors
static int dotProductOfVectors(Vector v1,
Vector v2)
{
// Traverse the vectors from the end because the
// ancestors starts from the root of the
// tree and root of the tree is present at the end
int size1 = v1.size();
int size2 = v2.size();
int i = size1 - 1;
int j = size2 - 1;
int answer = 0;
// Traverse the vectors side by side and storing
// answer
while (i >= 0 && j >= 0)
{
answer = answer + (v1.get(i) * v2.get(j));
i--;
j--;
}
// Return dot product
return answer;
}
// Utility function to calculate the dot product of
// the ancestors of the keys
static int dotProductOfAncestorsUtil(node root1,
node root2, int K1,
int K2)
{
// To store the ancestors of each key
Vector F1 = new Vector();
Vector F2 = new Vector();
// If both keys are present in the first tree
if (printAncestors(root1, K1, F1)
&& printAncestors(root1, K2, F2)) {
return dotProductOfVectors(F1, F2);
}
// If both keys are present in the second tree
else if (printAncestors(root2, K1, F1)
&& printAncestors(root2, K2, F2)) {
return dotProductOfVectors(F1, F2);
}
// If first key exists in first tree and
// the second key exists in second tree
else if (printAncestors(root1, K1, F1)
&& printAncestors(root2, K2, F2)) {
return dotProductOfVectors(F1, F2);
}
// Otherwise, first key exists in second tree and
// the second key exists in first tree
else {
if (printAncestors(root2, K1, F1)
&& printAncestors(root1, K2, F2)) {
return dotProductOfVectors(F1, F2);
}
}
// If either of the nodes doesn't exist
return 0;
}
static void dotProductOfAncestors(node root1,
node root2, int K1,
int K2)
{
// To store dot product of two vwctors
int dotProduct = dotProductOfAncestorsUtil(
root1, root2, K1, K2);
// Print dot product as the answer
System.out.println(dotProduct);
}
// Driver Code
public static void main(String[] args)
{
/* Construct the following binary trees
1 6
/ \ / \
2 3 7 8
/ \ / \
4 5 9 10
*/
// Given Tree 1
node root1 = newNode(1);
root1.left = newNode(2);
root1.right = newNode(3);
root1.left.left = newNode(4);
root1.left.right = newNode(5);
// Given Tree 2
node root2 = newNode(6);
root2.left = newNode(7);
root2.right = newNode(8);
root2.left.left = newNode(9);
root2.right.right = newNode(10);
// Given keys
int K1 = 4, K2 = 5;
// Function Call
dotProductOfAncestors(root1, root2, K1, K2);
}
}
// This code is contributed by Dharanendra L V Python3
# Python3 program for the above approach
# Structure of the tree node
class Node:
def __init__(self, x):
self.data = x
self.left = None
self.right = None
# Function to store the ancestors
# of the given key in the vector
def printAncestors(root, K, v):
# Global v
if (root == None):
return False
if (root.data == K):
return True
# If target is present in either left
# or right subtree of this node,
# then prthis node
if (printAncestors(root.left, K, v) or
printAncestors(root.right, K, v)):
v.append(root.data)
return True
return False
# Function to store the dot product
# of the vectors
def dotProductOfVectors(v1, v2):
# Global v1,v2
# Ancestors starts from the root of
# the tree and root of the tree is
# present at the end
size1 = len(v1)
size2 = len(v2)
i = size1 - 1
j = size2 - 1
answer = 0
# Traverse the vectors side by
# side and storing answer
while (i >= 0 and j >= 0):
answer = answer + (v1[i] * v2[j])
i -= 1
j -= 1
# Return dot product
return answer
# Utility function to calculate the dot product
# of the ancestors of the keys
def dotProductOfAncestorsUtil(root1, root2,
K1, K2):
# To store the ancestors of each key
F1, F2 = [], []
# If both keys are present in the first tree
if (printAncestors(root1, K1, F1) and
printAncestors(root1, K2, F2)):
return dotProductOfVectors(F1, F2)
# If both keys are present in the second tree
elif (printAncestors(root2, K1, F1) and
printAncestors(root2, K2, F2)):
return dotProductOfVectors(F1, F2)
# If first key exists in first tree and
# the second key exists in second tree
elif (printAncestors(root1, K1, F1) and
printAncestors(root2, K2, F2)):
return dotProductOfVectors(F1, F2)
# Otherwise, first key exists in second tree
# and the second key exists in first tree
else:
if (printAncestors(root2, K1, F1) and
printAncestors(root1, K2, F2)):
return dotProductOfVectors(F1, F2)
# If either of the nodes doesn't exist
return 0
def dotProductOfAncestors(root1, root2, K1, K2):
# To store dot product of two vwctors
dotProduct = dotProductOfAncestorsUtil(
root1, root2, K1, K2)
# Print dot product as the answer
print (dotProduct)
# Driver Code
if __name__ == '__main__':
v, v1, v2 = [], [], []
# Construct the following binary trees
# 1 6
# / \ / \
# 2 3 7 8
# / \ / \
# 4 5 9 10
# Given Tree 1
root1 = Node(1)
root1.left = Node(2)
root1.right = Node(3)
root1.left.left = Node(4)
root1.left.right = Node(5)
# Given Tree 2
root2 = Node(6)
root2.left = Node(7)
root2.right = Node(8)
root2.left.left = Node(9)
root2.right.right = Node(10)
# Given keys
K1, K2 = 4, 5
# Function Call
dotProductOfAncestors(root1, root2, K1, K2)
# This code is contributed by mohit kumar 29C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
// Structure of the tree node
public class node
{
public int data;
public node left, right;
}
// Utility function to create a new node
public static node newNode(int data)
{
node Node = new node();
Node.data = data;
Node.left = null;
Node.right = null;
return (Node);
}
// Function to store the ancestors
// of the given key in the vector
static bool printAncestors(node root, int K, List v)
{
// Base case
if (root == null)
return false;
if (root.data == K)
return true;
// If target is present in either left
// or right subtree of this node,
// then print this node
if (printAncestors(root.left, K, v)
|| printAncestors(root.right, K, v))
{
v.Add(root.data);
return true;
}
return false;
}
// Function to store the dot product of the vectors
static int dotProductOfVectors(List v1, List v2)
{
// Traverse the vectors from the end because the
// ancestors starts from the root of the
// tree and root of the tree is present at the end
int size1 = v1.Count;
int size2 = v2.Count;
int i = size1 - 1;
int j = size2 - 1;
int answer = 0;
// Traverse the vectors side by side and storing
// answer
while (i >= 0 && j >= 0)
{
answer = answer + (v1[i] * v2[j]);
i--;
j--;
}
// Return dot product
return answer;
}
// Utility function to calculate the dot product of
// the ancestors of the keys
static int dotProductOfAncestorsUtil(node root1,
node root2, int K1,
int K2)
{
// To store the ancestors of each key
List F1 = new List();
List F2 = new List();
// If both keys are present in the first tree
if (printAncestors(root1, K1, F1)
&& printAncestors(root1, K2, F2)) {
return dotProductOfVectors(F1, F2);
}
// If both keys are present in the second tree
else if (printAncestors(root2, K1, F1)
&& printAncestors(root2, K2, F2)) {
return dotProductOfVectors(F1, F2);
}
// If first key exists in first tree and
// the second key exists in second tree
else if (printAncestors(root1, K1, F1)
&& printAncestors(root2, K2, F2)) {
return dotProductOfVectors(F1, F2);
}
// Otherwise, first key exists in second tree and
// the second key exists in first tree
else {
if (printAncestors(root2, K1, F1)
&& printAncestors(root1, K2, F2)) {
return dotProductOfVectors(F1, F2);
}
}
// If either of the nodes doesn't exist
return 0;
}
static void dotProductOfAncestors(node root1,
node root2, int K1,
int K2)
{
// To store dot product of two vwctors
int dotProduct = dotProductOfAncestorsUtil(
root1, root2, K1, K2);
// Print dot product as the answer
Console.WriteLine(dotProduct);
}
static void Main() {
/* Construct the following binary trees
1 6
/ \ / \
2 3 7 8
/ \ / \
4 5 9 10
*/
// Given Tree 1
node root1 = newNode(1);
root1.left = newNode(2);
root1.right = newNode(3);
root1.left.left = newNode(4);
root1.left.right = newNode(5);
// Given Tree 2
node root2 = newNode(6);
root2.left = newNode(7);
root2.right = newNode(8);
root2.left.left = newNode(9);
root2.right.right = newNode(10);
// Given keys
int K1 = 4, K2 = 5;
// Function Call
dotProductOfAncestors(root1, root2, K1, K2);
}
}
// This code is contributed by divyeshrabadiya07. 输出：
5时间复杂度： O(N)，其中 N 是树中的节点数。
辅助空间： O(N)
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