N 元树中深度最大的最小值节点
给定一个由 N 个节点组成的树，任务是从根节点开始寻找深度最大的节点，取深度为零的根节点。如果有 1 个以上的最大深度节点，则找到具有最小值的节点。
例:
Input:
1
/ \
2 3
/ \
4 5
Output: 4
Explanation:
For this tree:
Height of Node 1 - 0,
Height of Node 2 - 1,
Height of Node 3 - 1,
Height of Node 4 - 2,
Height of Node 5 - 2\.
Hence, the nodes whose height is
maximum are 4 and 5, out of which
4 is minimum valued.
Input:
1
/
2
/
3
Output: 3
Explanation:
For this tree:
Height of Node 1 - 0,
Height of Node 2 - 1,
Height of Node 3 - 2
Hence, the node whose height
is maximum is 3.
方法:
- The idea is to use Depth First Search (DFS) on the tree. For each node, check the height of each node when we move down the tree.
- Check whether it is the largest so far, and if its height is equal to the maximum, then whether it is the minimum node.
- If so, update the maximum height and node values up to now accordingly.
下面是上述方法的实现:
c++
// C++ implementation of for
// the above problem
#include <bits/stdc++.h>
using namespace std;
#define MAX 100000
vector<int> graph[MAX + 1];
// To store the height of each node
int maxHeight, minNode;
// Function to perform dfs
void dfs(int node, int parent,
int h)
{
// Store the height of node
int height = h;
if (height > maxHeight) {
maxHeight = height;
minNode = node;
}
else if (height == maxHeight
&& minNode > node)
minNode = node;
for (int to : graph[node]) {
if (to == parent)
continue;
dfs(to, node, h + 1);
}
}
// Driver code
int main()
{
// Number of nodes
int N = 5;
// Edges of the tree
graph[1].push_back(2);
graph[1].push_back(3);
graph[2].push_back(4);
graph[2].push_back(5);
maxHeight = 0;
minNode = 1;
dfs(1, 1, 0);
cout << minNode << "\n";
return 0;
}
Java
// Java implementation of for
// the above problem
import java.util.*;
class GFG{
static final int MAX = 100000;
@SuppressWarnings("unchecked")
static Vector<Integer>[] graph = new Vector[MAX + 1];
// To store the height of each node
static int maxHeight, minNode;
// Function to perform dfs
static void dfs(int node, int parent, int h)
{
// Store the height of node
int height = h;
if (height > maxHeight)
{
maxHeight = height;
minNode = node;
}
else if (height == maxHeight &&
minNode > node)
minNode = node;
for(int to : graph[node])
{
if (to == parent)
continue;
dfs(to, node, h + 1);
}
}
// Driver code
public static void main(String[] args)
{
// Number of nodes
int N = 5;
for(int i = 0; i < graph.length; i++)
graph[i] = new Vector<Integer>();
// Edges of the tree
graph[1].add(2);
graph[1].add(3);
graph[2].add(4);
graph[2].add(5);
maxHeight = 0;
minNode = 1;
dfs(1, 1, 0);
System.out.print(minNode + "\n");
}
}
// This code is contributed by sapnasingh4991
python 3
# Python3 implementation of for
# the above problem
MAX = 100000
graph = [[] for i in range(MAX + 1)]
# To store the height of each node
maxHeight = 0
minNode = 0
# Function to perform dfs
def dfs(node, parent, h):
global minNode, maxHeight
# Store the height of node
height = h
if (height > maxHeight):
maxHeight = height
minNode = node
elif (height == maxHeight and
minNode > node):
minNode = node
for to in graph[node]:
if to == parent:
continue
dfs(to, node, h + 1)
# Driver code
if __name__=="__main__":
# Number of nodes
N = 5
# Edges of the tree
graph[1].append(2)
graph[1].append(3)
graph[2].append(4)
graph[2].append(5)
maxHeight = 0
minNode = 1
dfs(1, 1, 0)
print(minNode)
# This code is contributed by rutvik_56
c
// C# implementation of for
// the above problem
using System;
using System.Collections.Generic;
public class GFG{
static readonly int MAX = 100000;
static List<int>[] graph = new List<int>[MAX + 1];
// To store the height of each node
static int maxHeight, minNode;
// Function to perform dfs
static void dfs(int node, int parent, int h)
{
// Store the height of node
int height = h;
if (height > maxHeight)
{
maxHeight = height;
minNode = node;
}
else if (height == maxHeight &&
minNode > node)
minNode = node;
foreach(int to in graph[node])
{
if (to == parent)
continue;
dfs(to, node, h + 1);
}
}
// Driver code
public static void Main(String[] args)
{
for(int i = 0; i < graph.Length; i++)
graph[i] = new List<int>();
// Edges of the tree
graph[1].Add(2);
graph[1].Add(3);
graph[2].Add(4);
graph[2].Add(5);
maxHeight = 0;
minNode = 1;
dfs(1, 1, 0);
Console.Write(minNode + "\n");
}
}
// This code is contributed by shikhasingrajput
Javascript
<script>
// Javascript implementation of for the above problem
let MAX = 100000;
let graph = new Array(MAX + 1);
// To store the height of each node
let maxHeight, minNode;
// Function to perform dfs
function dfs(node, parent, h)
{
// Store the height of node
let height = h;
if (height > maxHeight)
{
maxHeight = height;
minNode = node;
}
else if (height == maxHeight &&
minNode > node)
minNode = node;
for(let to = 0; to < graph[node].length; to++)
{
if (graph[node][to] == parent)
continue;
dfs(graph[node][to], node, h + 1);
}
}
for(let i = 0; i < graph.length; i++)
graph[i] = [];
// Edges of the tree
graph[1].push(2);
graph[1].push(3);
graph[2].push(4);
graph[2].push(5);
maxHeight = 0;
minNode = 1;
dfs(1, 1, 0);
document.write(minNode + "</br>");
// This code is contributed by decode2207.
</script>
输出:
4
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