Dots binārs koks, uzdevums ir atrast koka augstumu. Koka augstums ir virsotņu skaits kokā no saknes līdz dziļākajam mezglam.

Piezīme: Tukša koka augstums ir 0 un koka ar vienu mezglu augstums ir 1 .

Binārā koka piemērs

Rekursīvi aprēķiniet augstumu pa kreisi un pa labi mezgla apakškokus un piešķir mezglam augstumu kā Maksimālais divu bērnu augums plus 1 . Plašāku informāciju skatiet tālāk pseido kodu un programmu.

Ilustrācija:

Apsveriet šādu koku:

Koka piemērs

maks.Dziļums('1') = maks(maksimālaisDziļums('2'), maxDepth('3')) + 1 = 2 + 1

jo rekursīvi
maxDepth('2') = max (maxDepth('4'), maxDepth('5')) + 1 = 1 + 1 un (jo gan '4', gan '5' augstums ir 1)
maxDepth('3') = 1

Lai īstenotu ideju, veiciet tālāk norādītās darbības.

• Rekursīvi veiciet meklēšanu pēc dziļuma.
• Ja koks ir tukšs, atgrieziet 0
• Pretējā gadījumā rīkojieties šādi
  • Rekursīvi iegūstiet kreisā apakškoka maksimālo dziļumu, t.i., izsauciet maxDepth(koks->kreisais apakškoks)
  • Rekursīvi iegūstiet īstā apakškoka maksimālo dziļumu, t.i., izsauciet maxDepth(koks->labais-apakškoks)
  • Iegūstiet maksimālo maksimālo dziļumu pa kreisi un pa labi apakškoki un pievienot 1 uz to pašreizējam mezglam.
    • 
• Atgriezties max_depth.

Zemāk ir aprakstīta iepriekš minētās pieejas īstenošana:

C++

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C

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Java

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Python3

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C#

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Javascript

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Height of tree is 3>

Laika sarežģītība: O(N) (Lūdzu, skatiet ziņu vietnē Koku šķērsošana sīkākai informācijai)
Palīgtelpa: O(N) rekursīvās steka dēļ.

Atrodiet koka maksimālo dziļumu vai augstumu, izmantojot Līmeņa pasūtījumu izbraukšana :

Dariet Līmeņa pasūtījumu izbraukšana , vienlaikus pievienojot mezglus katrā līmenī Lai īstenotu ideju, veiciet tālāk norādītās darbības.

• Šķērsojiet koku līmenī, šķērsojot, sākot no sakne .
  • Inicializējiet tukšu rindu J , mainīgais dziļums un spiediet sakne , pēc tam nospiediet null iekšā J .
  • Palaidiet kamēr cilpa līdz J nav tukšs.
    • Uzglabājiet priekšējo elementu J un Izvelciet priekšējo elementu.
    • Ja priekšpuse J ir NULL tad palielinājums dziļums pa vienam un, ja rinda nav tukša, tad spiediet NULL iekšā J .
    • Citādi, ja elements nav NULL tad pārbaudiet to pa kreisi un pa labi bērni un ja tādi nav NULL iespiediet tos J .
• Atgriezties dziļums .

Zemāk ir aprakstīta iepriekš minētās pieejas īstenošana:

C++

#include> #include> using> namespace> std;> // A Tree node> struct> Node {> >int> key;> >struct> Node *left, *right;> };> // Utility function to create a new node> Node* newNode(>int> key)> {> >Node* temp =>new> Node;> >temp->taustiņš = atslēga;> >temp->pa kreisi = temp->right = NULL;> >return> (temp);> }> /*Function to find the height(depth) of the tree*/> int> height(>struct> Node* root)> {> >// Initialising a variable to count the> >// height of tree> >int> depth = 0;> >queue q;> >// Pushing first level element along with NULL> >q.push(root);> >q.push(NULL);> >while> (!q.empty()) {> >Node* temp = q.front();> >q.pop();> >// When NULL encountered, increment the value> >if> (temp == NULL) {> >depth++;> >}> >// If NULL not encountered, keep moving> >if> (temp != NULL) {> >if> (temp->pa kreisi) {> >q.push(temp->pa kreisi);> >}> >if> (temp->pa labi) {> >q.push(temp->pa labi);> >}> >}> >// If queue still have elements left,> >// push NULL again to the queue.> >else> if> (!q.empty()) {> >q.push(NULL);> >}> >}> >return> depth;> }> // Driver program> int> main()> {> >// Let us create Binary Tree shown in above example> >Node* root = newNode(1);> >root->pa kreisi = newNode(2);> >root->pa labi = newNode(3);> >root->pa kreisi->pa kreisi = newNode(4);> >root->pa kreisi->pa labi = newNode(5);> >cout <<>'Height(Depth) of tree is: '> << height(root);> }>

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Java

// Java program for above approach> import> java.util.LinkedList;> import> java.util.Queue;> class> GFG {> >// A tree node structure> >static> class> Node {> >int> key;> >Node left;> >Node right;> >}> >// Utility function to create> >// a new node> >static> Node newNode(>int> key)> >{> >Node temp =>new> Node();> >temp.key = key;> >temp.left = temp.right =>null>;> >return> temp;> >}> >/*Function to find the height(depth) of the tree*/> >public> static> int> height(Node root)> >{> >// Initialising a variable to count the> >// height of tree> >int> depth =>0>;> >Queue q =>new> LinkedList();> >// Pushing first level element along with null> >q.add(root);> >q.add(>null>);> >while> (!q.isEmpty()) {> >Node temp = q.peek();> >q.remove();> >// When null encountered, increment the value> >if> (temp ==>null>) {> >depth++;> >}> >// If null not encountered, keep moving> >if> (temp !=>null>) {> >if> (temp.left !=>null>) {> >q.add(temp.left);> >}> >if> (temp.right !=>null>) {> >q.add(temp.right);> >}> >}> >// If queue still have elements left,> >// push null again to the queue.> >else> if> (!q.isEmpty()) {> >q.add(>null>);> >}> >}> >return> depth;> >}> >// Driver Code> >public> static> void> main(String args[])> >{> >Node root = newNode(>1>);> >root.left = newNode(>2>);> >root.right = newNode(>3>);> >root.left.left = newNode(>4>);> >root.left.right = newNode(>5>);> >System.out.println(>'Height(Depth) of tree is: '> >+ height(root));> >}> }> // This code is contributed by jana_sayantan.>

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Python3

# Python code to implement the approach> # A Tree node> class> Node:> >def> __init__(>self>):> >self>.key>=> 0> >self>.left,>self>.right>=> None>,>None> # Utility function to create a new node> def> newNode(key):> >temp>=> Node()> >temp.key>=> key> >temp.left, temp.right>=> None>,>None> >return> temp> # Function to find the height(depth) of the tree> def> height(root):> ># Initialising a variable to count the> ># height of tree> >depth>=> 0> >q>=> []> ># appending first level element along with None> >q.append(root)> >q.append(>None>)> >while>(>len>(q)>>>):> >temp>=> q[>0>]> >q>=> q[>1>:]> ># When None encountered, increment the value> >if>(temp>=>=> None>):> >depth>+>=> 1> ># If None not encountered, keep moving> >if>(temp !>=> None>):> >if>(temp.left):> >q.append(temp.left)> >if>(temp.right):> >q.append(temp.right)> ># If queue still have elements left,> ># append None again to the queue.> >elif>(>len>(q)>>>):> >q.append(>None>)> >return> depth> # Driver program> # Let us create Binary Tree shown in above example> root>=> newNode(>1>)> root.left>=> newNode(>2>)> root.right>=> newNode(>3>)> root.left.left>=> newNode(>4>)> root.left.right>=> newNode(>5>)> print>(f>'Height(Depth) of tree is: {height(root)}'>)> # This code is contributed by shinjanpatra>

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C#

// C# Program to find the Maximum Depth or Height of Binary Tree> using> System;> using> System.Collections.Generic;> // A Tree node> public> class> Node {> >public> int> data;> >public> Node left, right;> >public> Node(>int> item)> >{> >data = item;> >left =>null>;> >right =>null>;> >}> }> public> class> BinaryTree {> >Node root;> >// Function to find the height(depth) of the tree> >int> height()> >{> >// Initialising a variable to count the> >// height of tree> >int> depth = 0;> >Queue q =>new> Queue();> >// Pushing first level element along with NULL> >q.Enqueue(root);> >q.Enqueue(>null>);> >while> (q.Count != 0) {> >Node temp = q.Dequeue();> >// When NULL encountered, increment the value> >if> (temp ==>null>)> >depth++;> >// If NULL not encountered, keep moving> >if> (temp !=>null>) {> >if> (temp.left !=>null>) {> >q.Enqueue(temp.left);> >}> >if> (temp.right !=>null>) {> >q.Enqueue(temp.right);> >}> >}> >// If queue still have elements left,> >// push NULL again to the queue> >else> if> (q.Count != 0) {> >q.Enqueue(>null>);> >}> >}> >return> depth;> >}> >// Driver program> >public> static> void> Main()> >{> >// Let us create Binary Tree shown in above example> >BinaryTree tree =>new> BinaryTree();> >tree.root =>new> Node(1);> >tree.root.left =>new> Node(2);> >tree.root.right =>new> Node(3);> >tree.root.left.left =>new> Node(4);> >tree.root.left.right =>new> Node(5);> >Console.WriteLine(>'Height(Depth) of tree is: '> >+ tree.height());> >}> }> // This code is contributed by Yash Agarwal(yashagarwal2852002)>

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Javascript

> // JavaScript code to implement the approach> // A Tree node> class Node{> >constructor(){> >this>.key = 0> >this>.left =>null> >this>.right =>null> >}> }> // Utility function to create a new node> function> newNode(key){> >let temp =>new> Node()> >temp.key = key> >temp.left =>null> >temp.right =>null> >return> temp> }> // Function to find the height(depth) of the tree> function> height(root){> >// Initialising a variable to count the> >// height of tree> >let depth = 0> >let q = []> > >// pushing first level element along with null> >q.push(root)> >q.push(>null>)> >while>(q.length>0){> >let temp = q.shift()> > >// When null encountered, increment the value> >if>(temp ==>null>)> >depth += 1> > >// If null not encountered, keep moving> >if>(temp !=>null>){> >if>(temp.left)> >q.push(temp.left)> > >if>(temp.right)> >q.push(temp.right)> >}> > >// If queue still have elements left,> >// push null again to the queue.> >else> if>(q.length>0)> >q.push(>null>)> >}> >return> depth> }> // Driver program> // Let us create Binary Tree shown in above example> let root = newNode(1)> root.left = newNode(2)> root.right = newNode(3)> root.left.left = newNode(4)> root.left.right = newNode(5)> document.write(`Height(Depth) of tree is: ${height(root)}`,>''>)> // This code is contributed by shinjanpatra> >

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Izvade

Height(Depth) of tree is: 3>

Laika sarežģītība: O(N)
Palīgtelpa: O(N)

Vēl viena metode augstuma noteikšanai, izmantojot Līmeņa pasūtījumu izbraukšana :

Šajā metodē tiek izmantots arī jēdziens Level Order Traversal, taču mēs rindā nepievienosim nulli. Vienkārši palieliniet skaitītājs kad līmenis paaugstinās un iebīdiet rindā esošā mezgla bērnus, pēc tam noņemiet visus mezglus no pašreizējā līmeņa rindas.

c# vārdnīca

C++

// C++ program for above approach> #include> using> namespace> std;> // A Tree node> struct> Node {> >int> key;> >struct> Node *left, *right;> };> // Utility function to create a new node> Node* newNode(>int> key)> {> >Node* temp =>new> Node;> >temp->taustiņš = atslēga;> >temp->pa kreisi = temp->right = NULL;> >return> (temp);> }> /*Function to find the height(depth) of the tree*/> int> height(Node* root)> {> >// Initialising a variable to count the> >// height of tree> >queue q;> >q.push(root);> >int> height = 0;> >while> (!q.empty()) {> >int> size = q.size();> >for> (>int> i = 0; i Node* temp = q.front(); q.pop(); if (temp->pa kreisi != NULL) { q.push(temp->left); } if (temp->right != NULL) { q.push(temp->right); } } augstums++; } atgriešanas augstums; } // Draivera programma int main() { // Izveidosim bināro koku, kas parādīts augstāk esošajā piemērā Node* root = newNode(1); sakne->pa kreisi = newNode(2); sakne->pa labi = newNode(3); sakne->pa kreisi->pa kreisi = newNode(4); sakne->pa kreisi->pa labi = newNode(5); cout<< 'Height(Depth) of tree is: ' << height(root); } // This code is contributed by Abhijeet Kumar(abhijeet19403)>

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Java

// Java program for above approach> import> java.util.LinkedList;> import> java.util.Queue;> class> GFG {> >// A tree node structure> >static> class> Node {> >int> key;> >Node left;> >Node right;> >}> >// Utility function to create> >// a new node> >static> Node newNode(>int> key)> >{> >Node temp =>new> Node();> >temp.key = key;> >temp.left = temp.right =>null>;> >return> temp;> >}> >/*Function to find the height(depth) of the tree*/> >public> static> int> height(Node root)> >{> >// Initialising a variable to count the> >// height of tree> >Queue q =>new> LinkedList();> >q.add(root);> >int> height =>0>;> >while> (!q.isEmpty()) {> >int> size = q.size();> >for> (>int> i =>0>; i Node temp = q.poll(); if (temp.left != null) { q.add(temp.left); } if (temp.right != null) { q.add(temp.right); } } height++; } return height; } // Driver Code public static void main(String args[]) { Node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); System.out.println('Height(Depth) of tree is: ' + height(root)); } }>

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Python3

# Python3 program to find the height of a tree> > # A binary tree node> class> Node:> > ># Constructor to create a new node> >def> __init__(>self>, data):> >self>.key>=> data> >self>.left>=> None> >self>.right>=> None> > # Function to find height of tree> def> height(root):> ># Base Case> >if> root>is> None>:> >return> 0> > ># Create an empty queue for level order traversal> >q>=> []> > ># Enqueue Root and initialize height> >q.append(root)> >height>=> 0> > ># Loop while queue is not empty> >while> q:> > ># nodeCount (queue size) indicates number of nodes> ># at current level> >nodeCount>=> len>(q)> > ># Dequeue all nodes of current level and Enqueue all> ># nodes of next level> >while> nodeCount>>>:> >node>=> q.pop(>0>)> >if> node.left>is> not> None>:> >q.append(node.left)> >if> node.right>is> not> None>:> >q.append(node.right)> >nodeCount>->=> 1> >height>+>=> 1> > >return> height> > # Driver Code> root>=> Node(>1>)> root.left>=> Node(>2>)> root.right>=> Node(>3>)> root.left.left>=> Node(>4>)> root.left.right>=> Node(>5>)> > print>(>'Height(Depth) of tree is'>, height(root))>

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C#

using> System;> using> System.Collections.Generic;> class> GFG {> >// A Tree node> >class> Node {> >public> int> key;> >public> Node left, right;> >public> Node(>int> key)> >{> >this>.key=key;> >this>.left=>this>.right=>null>;> >}> >}> >// Utility function to create a new node> >/*Node newNode(int key)> >{> >Node* temp = new Node;> >temp.key = key;> >temp.left = temp.right = NULL;> >return (temp);> >}*/> >/*Function to find the height(depth) of the tree*/> >static> int> height(Node root)> >{> >// Initialising a variable to count the> >// height of tree> >Queue q=>new> Queue();> >q.Enqueue(root);> >int> height = 0;> >while> (q.Count>0) {> >int> size = q.Count;> >for> (>int> i = 0; i Node temp = q.Peek(); q.Dequeue(); if (temp.left != null) { q.Enqueue(temp.left); } if (temp.right != null) { q.Enqueue(temp.right); } } height++; } return height; } // Driver program public static void Main() { // Let us create Binary Tree shown in above example Node root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); Console.Write('Height(Depth) of tree is: ' + height(root)); } } // This code is contributed by poojaagarwal2.>

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Javascript

// JavaScript program for above approach> // a tree node> class Node{> >constructor(key){> >this>.key = key;> >this>.left =>this>.right =>null>;> >}> }> // utility function to create a new node> function> newNode(key){> >return> new> Node(key);> }> // function to find the height of the tree> function> height(root){> >// initialising a variable to count the> >// height of tree> >let q = [];> >q.push(root);> >let height = 0;> >while>(q.length>0){> >let size = q.length;> >for>(let i = 0; i let temp = q.shift(); if(temp.left != null){ q.push(temp.left); } if(temp.right != null){ q.push(temp.right); } } height++; } return height; } // driver code let root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); document.write('Height(Depth) of tree is: ' + height(root)); // this code is contributed by Kirti Agarwal(kirtiagarwal23121999)>

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Izvade

Height(Depth) of tree is: 3>

Laika sarežģītība: O(N)
Palīgtelpa: O(N)