给定二叉树v1和v2的两个节点，任务是检查两个节点是否在树中的同一路径上。
例子：
Input: v1 = 1, v2 = 5
1
/ | \
2 3 4
/ | \
5 6 7
Output: Yes
Explanation:
Both nodes 1 and 5
lie in the path 1 -> 2 -> 5.
Input: v1 = 2, v2 = 6
1
/ | \
2 3 4
/ | \
5 6 7
Output: NO
DFS方法：有关DFS方法，请参阅检查树中两个节点是否在同一路径上。
LCA方法：这个想法是使用最低共同祖先。找到给定顶点v1和v2的LCA。如果LCA等于给定两个顶点中的任何一个，则打印“是” 。否则，打印No。
下面是上述方法的实现：
C++
// C++ program to check if two nodes
// are on same path in a tree without
// using any extra space
#include
using namespace std;
// Function to filter
// the return Values
int filter(int x, int y, int z)
{
if (x != -1 && y != -1) {
return z;
}
return x == -1 ? y : x;
}
// Utility function to check if nodes
// are on same path or not
int samePathUtil(int mtrx[][7], int vrtx,
int v1, int v2, int i)
{
int ans = -1;
// Condition to check
// if any vertex
// is equal to given two
// vertex or not
if (i == v1 || i == v2)
return i;
for (int j = 0; j < vrtx; j++) {
// Check if the current
// position has 1
if (mtrx[i][j] == 1) {
// Recursive call
ans
= filter(ans,
samePathUtil(mtrx,
vrtx,
v1,
v2,
j),
i);
}
}
// Return LCA
return ans;
}
// Function to check if nodes
// lies on same path or not
bool isVertexAtSamePath(int mtrx[][7],
int vrtx, int v1,
int v2, int i)
{
int lca = samePathUtil(mtrx,
vrtx, v1 - 1,
v2 - 1, i);
if (lca == v1 - 1 || lca == v2 - 1)
return true;
return false;
}
// Driver Program
int main()
{
int vrtx = 7, edge = 6;
int mtrx[7][7] = {
{ 0, 1, 1, 1, 0, 0, 0 },
{ 0, 0, 0, 0, 1, 0, 0 },
{ 0, 0, 0, 0, 0, 1, 0 },
{ 0, 0, 0, 0, 0, 0, 1 },
{ 0, 0, 0, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0, 0, 0 }
};
int v1 = 1, v2 = 5;
if (isVertexAtSamePath(mtrx,
vrtx, v1,
v2, 0))
cout << "Yes";
else
cout << "No";
return 0;
} Java
// Java program to check if two nodes
// are on same path in a tree without
// using any extra space
class GFG{
// Function to filter
// the return Values
static int filter(int x, int y, int z)
{
if (x != -1 && y != -1)
{
return z;
}
return x == -1 ? y : x;
}
// Utility function to check if nodes
// are on same path or not
static int samePathUtil(int mtrx[][], int vrtx,
int v1, int v2, int i)
{
int ans = -1;
// Condition to check
// if any vertex
// is equal to given two
// vertex or not
if (i == v1 || i == v2)
return i;
for(int j = 0; j < vrtx; j++)
{
// Check if the current
// position has 1
if (mtrx[i][j] == 1)
{
// Recursive call
ans = filter(ans, samePathUtil(
mtrx, vrtx, v1,
v2, j), i);
}
}
// Return LCA
return ans;
}
// Function to check if nodes
// lies on same path or not
static boolean isVertexAtSamePath(int mtrx[][],
int vrtx, int v1,
int v2, int i)
{
int lca = samePathUtil(mtrx, vrtx, v1 - 1,
v2 - 1, i);
if (lca == v1 - 1 || lca == v2 - 1)
return true;
return false;
}
// Driver code
public static void main(String[] args)
{
int vrtx = 7;
int mtrx[][] = { { 0, 1, 1, 1, 0, 0, 0 },
{ 0, 0, 0, 0, 1, 0, 0 },
{ 0, 0, 0, 0, 0, 1, 0 },
{ 0, 0, 0, 0, 0, 0, 1 },
{ 0, 0, 0, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0, 0, 0 } };
int v1 = 1, v2 = 5;
if (isVertexAtSamePath(mtrx, vrtx,
v1, v2, 0))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by Rajput-JiPython3
# Python3 program to check if two nodes
# are on same path in a tree without
# using any extra space
# Function to filter
# the return Values
def filter(x, y, z):
if (x != -1 and y != -1):
return z
return y if x == -1 else x
# Utility function to check if nodes
# are on same path or not
def samePathUtil(mtrx, vrtx, v1, v2, i):
ans = -1
# Condition to check
# if any vertex
# is equal to given two
# vertex or not
if (i == v1 or i == v2):
return i
for j in range(0, vrtx):
# Check if the current
# position has 1
if (mtrx[i][j] == 1):
# Recursive call
ans = filter(ans,
samePathUtil(mtrx, vrtx,
v1, v2, j), i)
# Return LCA
return ans
# Function to check if nodes
# lies on same path or not
def isVertexAtSamePath(mtrx, vrtx, v1, v2, i):
lca = samePathUtil(mtrx, vrtx, v1 - 1,
v2 - 1, i)
if (lca == v1 - 1 or lca == v2 - 1):
return True
return False
# Driver code
vrtx = 7
edge = 6
mtrx = [ [ 0, 1, 1, 1, 0, 0, 0 ] ,
[ 0, 0, 0, 0, 1, 0, 0 ],
[ 0, 0, 0, 0, 0, 1, 0 ],
[ 0, 0, 0, 0, 0, 0, 1 ],
[ 0, 0, 0, 0, 0, 0, 0 ],
[ 0, 0, 0, 0, 0, 0, 0 ],
[ 0, 0, 0, 0, 0, 0, 0 ] ]
v1 = 1
v2 = 5
if (isVertexAtSamePath(mtrx, vrtx, v1, v2, 0)):
print("Yes")
else:
print("No")
# This code is contributed by sanjoy_62C#
// C# program to check if two nodes
// are on same path in a tree without
// using any extra space
using System;
class GFG{
// Function to filter
// the return Values
static int filter(int x, int y, int z)
{
if (x != -1 && y != -1)
{
return z;
}
return x == -1 ? y : x;
}
// Utility function to check if nodes
// are on same path or not
static int samePathUtil(int [,]mtrx, int vrtx,
int v1, int v2, int i)
{
int ans = -1;
// Condition to check
// if any vertex
// is equal to given two
// vertex or not
if (i == v1 || i == v2)
return i;
for(int j = 0; j < vrtx; j++)
{
// Check if the current
// position has 1
if (mtrx[i,j] == 1)
{
// Recursive call
ans = filter(ans, samePathUtil(
mtrx, vrtx, v1,
v2, j), i);
}
}
// Return LCA
return ans;
}
// Function to check if nodes
// lies on same path or not
static bool isVertexAtSamePath(int [,]mtrx,
int vrtx, int v1,
int v2, int i)
{
int lca = samePathUtil(mtrx, vrtx, v1 - 1,
v2 - 1, i);
if (lca == v1 - 1 || lca == v2 - 1)
return true;
return false;
}
// Driver code
public static void Main(String[] args)
{
int vrtx = 7;
int [,]mtrx = { { 0, 1, 1, 1, 0, 0, 0 },
{ 0, 0, 0, 0, 1, 0, 0 },
{ 0, 0, 0, 0, 0, 1, 0 },
{ 0, 0, 0, 0, 0, 0, 1 },
{ 0, 0, 0, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0, 0, 0 } };
int v1 = 1, v2 = 5;
if (isVertexAtSamePath(mtrx, vrtx,
v1, v2, 0))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by sapnasingh4991输出：
Yes
时间复杂度： O(N)
辅助空间： O(1)
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