Dado un árbol binario, escriba una función que devuelva el tamaño del subárbol más grande que también es un árbol de búsqueda binaria (BST). Si el árbol binario completo es BST, devuelve el tamaño de todo el árbol.
Ejemplos:
Input:
5
/ \
2 4
/ \
1 3
Output: 3
The following subtree is the
maximum size BST subtree
2
/ \
1 3
Input:
50
/ \
30 60
/ \ / \
5 20 45 70
/ \
65 80
Output: 5
The following subtree is the
maximum size BST subtree
60
/ \
45 70
/ \
65 80
Recomendado: Resuelva primero en » PRÁCTICA «, antes de pasar a la solución.
Hemos discutido dos métodos en la publicación a continuación.
Encuentre el subárbol BST más grande en un árbol binario dado | Conjunto 1
En esta publicación, se analiza una solución O(n) diferente. Esta solución es más simple que las soluciones discutidas anteriormente y funciona en tiempo O(n).
La idea se basa en el método 3 de verificar si un árbol binario es un artículo BST .
Un árbol es BST si lo siguiente es cierto para cada Node x.
- El valor más grande en el subárbol izquierdo (de x) es más pequeño que el valor de x.
- El valor más pequeño en el subárbol derecho (de x) es mayor que el valor de x.
Atravieso un árbol de manera de abajo hacia arriba. Para cada Node atravesado, devolvemos valores máximos y mínimos en el subárbol enraizado con él. Si algún Node sigue las propiedades anteriores y el tamaño de
C++14
// C++ program to find largest BST in a
// Binary Tree.
#include <bits/stdc++.h>
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);
}
// Information to be returned by every
// node in bottom up traversal.
struct Info
{
int sz; // Size of subtree
int max; // Min value in subtree
int min; // Max value in subtree
int ans; // Size of largest BST which
// is subtree of current node
bool isBST; // If subtree is BST
};
// Returns Information about subtree. The
// Information also includes size of largest
// subtree which is a BST.
Info largestBSTBT(Node* root)
{
// Base cases : When tree is empty or it has
// one child.
if (root == NULL)
return {0, INT_MIN, INT_MAX, 0, true};
if (root->left == NULL && root->right == NULL)
return {1, root->data, root->data, 1, true};
// Recur for left subtree and right subtrees
Info l = largestBSTBT(root->left);
Info r = largestBSTBT(root->right);
// Create a return variable and initialize its
// size.
Info ret;
ret.sz = (1 + l.sz + r.sz);
// If whole tree rooted under current root is
// BST.
if (l.isBST && r.isBST && l.max < root->data &&
r.min > root->data)
{
ret.min = min(l.min, min(r.min, root->data));
ret.max = max(r.max, max(l.max, root->data));
// Update answer for tree rooted under
// current 'root'
ret.ans = max(l.ans, r.ans) + 1;
ret.isBST = true;
return ret;
}
// If whole tree is not BST, return maximum
// of left and right subtrees
ret.ans = max(l.ans, r.ans);
ret.isBST = false;
return ret;
}
/* Driver program to test above functions*/
int main()
{
/* Let us construct the following Tree
60
/ \
65 70
/
50 */
struct Node *root = newNode(60);
root->left = newNode(65);
root->right = newNode(70);
root->left->left = newNode(50);
printf(" Size of the largest BST is %d\n",
largestBSTBT(root).ans);
return 0;
}
// This code is contributed by Vivek Garg in a
// comment on below set 1.
// www.geeksforgeeks.org/find-the-largest-subtree-in-a-tree-that-is-also-a-bst/
// This code is updated by Naman Sharma
Java
// Java program to find largest BST in a Binary Tree.
/* A binary tree node has data, pointer to left child and a
* pointer to right child */
class Node {
int data;
Node left, right;
public Node(final int d) { data = d; }
}
class GFG {
public static void main(String[] args)
{
/* Let us construct the following Tree
60
/ \
65 70
/
50 */
final Node node1 = new Node(60);
node1.left = new Node(65);
node1.right = new Node(70);
node1.left.left = new Node(50);
System.out.print("Size of the largest BST is "
+ Solution.largestBst(node1)
+ "\n");
}
}
class Solution {
static int MAX = Integer.MAX_VALUE;
static int MIN = Integer.MIN_VALUE;
static class nodeInfo {
int size; // Size of subtree
int max; // Min value in subtree
int min; // Max value in subtree
boolean isBST; // If subtree is BST
nodeInfo() {}
nodeInfo(int size, int max, int min, boolean isBST)
{
this.size = size;
this.max = max;
this.min = min;
this.isBST = isBST;
}
}
static nodeInfo largestBST(Node root)
{
// Base case : When the current subtree is empty or
// the node corresponds to a null.
if (root == null) {
return new nodeInfo(0, Integer.MIN_VALUE,
Integer.MAX_VALUE, true);
}
// We will here do the postorder traversal since we
// want our left and right subtrees to be computed
// first.
// Recur for left subtree and right subtrees
nodeInfo left = largestBST(root.left);
nodeInfo right = largestBST(root.right);
// Create a new nodeInfo variable to store info
// about the current node.
nodeInfo returnInfo = new nodeInfo();
returnInfo.min = Math.min(
Math.min(left.min, right.min), root.data);
returnInfo.max = Math.max(
Math.max(left.max, right.max), root.data);
returnInfo.isBST = left.isBST && right.isBST
&& root.data > left.max
&& root.data < right.min;
/*
If suppose the left and right subtrees of the
current node are BST and the current node value is
greater than the maximum value in the left subtree
and also the current node value is smaller that the
minimum value in the right subtree (Basic conditions
for the formation of BST) then our whole subtree
with the root as current root can be consider as a
BST. Hence the size of the BST will become size of
left BST subtree + size of right BST subtree +
1(adding current node).
Else we will consider the largest of the left BST or
the right BST.
*/
if (returnInfo.isBST)
returnInfo.size = left.size + right.size + 1;
else
returnInfo.size
= Math.max(left.size, right.size);
return returnInfo;
}
// Return the size of the largest sub-tree which is also
// a BST
static int largestBst(Node root)
{
return largestBST(root).size;
}
}
// This code is contributed by Andrei Sljusar
// This code is updated by Utkarsh Saxena.
Python3
# Python program to find largest
# BST in a Binary Tree.
INT_MIN = -2147483648
INT_MAX = 2147483647
# Helper function that allocates a new
# node with the given data and None left
# and right pointers.
class newNode:
# Constructor to create a new node
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Returns Information about subtree. The
# Information also includes size of largest
# subtree which is a BST
def largestBSTBT(root):
# Base cases : When tree is empty or it has
# one child.
if (root == None):
return 0, INT_MIN, INT_MAX, 0, True
if (root.left == None and root.right == None) :
return 1, root.data, root.data, 1, True
# Recur for left subtree and right subtrees
l = largestBSTBT(root.left)
r = largestBSTBT(root.right)
# Create a return variable and initialize its
# size.
ret = [0, 0, 0, 0, 0]
ret[0] = (1 + l[0] + r[0])
# If whole tree rooted under current root is
# BST.
if (l[4] and r[4] and l[1] <
root.data and r[2] > root.data) :
ret[2] = min(l[2], min(r[2], root.data))
ret[1] = max(r[1], max(l[1], root.data))
# Update answer for tree rooted under
# current 'root'
ret[3] = max(l[3], r[3]) + 1;
ret[4] = True
return ret
# If whole tree is not BST, return maximum
# of left and right subtrees
ret[3] = max(l[3], r[3])
ret[4] = False
return ret
# Driver Code
if __name__ == '__main__':
"""Let us construct the following Tree
60
/ \
65 70
/
50 """
root = newNode(60)
root.left = newNode(65)
root.right = newNode(70)
root.left.left = newNode(50)
print("Size of the largest BST is",
largestBSTBT(root)[3])
# This code is contributed
# Shubham Singh(SHUBHAMSINGH10)
# Naman Sharma (NAMANSHARMA1805)
C#
// C# program to find largest BST in a Binary Tree.
/* A binary tree node has data, pointer to left child and a
* pointer to right child */
using System;
public
class Node {
public
int data;
public
Node left,
right;
public Node(int d) { data = d; }
}
public class GFG {
public static void Main(String[] args)
{
/*
* Let us construct the following Tree 60 / \ 65 70
* / 50
*/
Node node1 = new Node(60);
node1.left = new Node(65);
node1.right = new Node(70);
node1.left.left = new Node(50);
Console.Write("Size of the largest BST is "
+ Solution.largestBst(node1) + "\n");
}
}
public
class Solution {
static int MAX = int.MaxValue;
static int MIN = int.MinValue;
public
class nodeInfo {
public int size; // Size of subtree
public int max; // Min value in subtree
public int min; // Max value in subtree
public bool isBST; // If subtree is BST
public
nodeInfo()
{
}
public
nodeInfo(int size, int max, int min, bool isBST)
{
this.size = size;
this.max = max;
this.min = min;
this.isBST = isBST;
}
}
public
static nodeInfo
largestBST(Node root)
{
// Base cases : When tree is empty or it has one
// child.
if (root == null) {
return new nodeInfo(0, MIN, MAX, true);
}
if (root.left == null && root.right == null) {
return new nodeInfo(1, root.data, root.data,
true);
}
// Recur for left subtree and right subtrees
nodeInfo left = largestBST(root.left);
nodeInfo right = largestBST(root.right);
// Create a return variable and initialize its size.
nodeInfo returnInfo = new nodeInfo();
// If whole tree rooted under current root is BST.
if (left.isBST && right.isBST
&& left.max < root.data
&& right.min > root.data) {
returnInfo.min = Math.Min(
Math.Min(left.min, right.min), root.data);
returnInfo.max = Math.Max(
Math.Max(left.max, right.max), root.data);
// Update answer for tree rooted under current
// 'root'
returnInfo.size = Math.Max(left.size, right.size) + 1;
returnInfo.isBST = true;
return returnInfo;
}
// If whole tree is not BST, return maximum of left
// and right subtrees
returnInfo.size = Math.Max(left.size, right.size);
returnInfo.isBST = false;
return returnInfo;
}
// Return the size of the largest sub-tree which is also
// a BST
public
static int
largestBst(Node root)
{
return largestBST(root).size;
}
}
// This code is contributed by Rajput-Ji
//This code is updated by Naman Sharma
Javascript
<script>
// javascript program to find largest BST in a Binary Tree.
/* A binary tree node has data, pointer to left child and a pointer to right child */
class Node {
constructor(val) {
this.data = val;
this.left = null;
this.right = null;
}
}
var MAX = Number.MAX_VALUE;
var MIN = Number.MIN_VALUE;
class nodeInfo {
constructor(size , max , min, isBST) {
this.size = size;
this.max = max;
this.min = min;
this.isBST = isBST;
}
}
function largestBST(root) {
// Base cases : When tree is empty or it has one child.
if (root == null) {
return new nodeInfo(0, MIN, MAX, true);
}
if (root.left == null && root.right == null) {
return new nodeInfo(1, root.data, root.data, true);
}
// Recur for left subtree and right subtrees
var left = largestBST(root.left);
var right = largestBST(root.right);
// Create a return variable and initialize its size.
var returnInfo = new nodeInfo();
// If whole tree rooted under current root is BST.
if (left.isBST && right.isBST && left.max < root.data && right.min > root.data) {
returnInfo.min = Math.min(Math.min(left.min, right.min), root.data);
returnInfo.max = Math.max(Math.max(left.max, right.max), root.data);
// Update answer for tree rooted under current 'root'
returnInfo.size = math.max(left.size, right.size) + 1;
returnInfo.isBST = true;
return returnInfo;
}
// If whole tree is not BST, return maximum of left and right subtrees
returnInfo.size = Math.max(left.size, right.size);
returnInfo.isBST = false;
return returnInfo;
}
// Return the size of the largest sub-tree which is also a BST
function largestBst(root) {
return largestBST(root).size;
}
/*
* Let us construct the following Tree
60
/ \
65 70
/
50
*/
var node1 = new Node(60);
node1.left = new Node(65);
node1.right = new Node(70);
node1.left.left = new Node(50);
document.write("Size of the largest BST is " + largestBst(node1) + "\n");
// This code is contributed by Rajput-Ji and Naman Sharma
</script>
Producción
Size of the largest BST is 2
Complejidad de tiempo : O(n)
Complejidad espacial : O (n) Para la pila de llamadas desde que se usa la recursividad
Este artículo es una contribución de Utkarsh Saxena . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA