How to Draw Tree From Preorder Traversal
Construct Tree from given Inorder and Preorder traversals
Let us consider the below traversals:
Inorder sequence: D B E A F C
Preorder sequence: A B D E C F
In a Preorder sequence, leftmost element is the root of the tree. So we know 'A' is root for given sequences. By searching 'A' in Inorder sequence, we can find out all elements on left side of 'A' are in left subtree and elements on right are in right subtree. So we know below structure now.
A / / D B E F C
We recursively follow above steps and get the following tree.
A / / B C / / / / D E F
Algorithm: buildTree()
1) Pick an element from Preorder. Increment a Preorder Index Variable (preIndex in below code) to pick next element in next recursive call.
2) Create a new tree node tNode with the data as picked element.
3) Find the picked element's index in Inorder. Let the index be inIndex.
4) Call buildTree for elements before inIndex and make the built tree as left subtree of tNode.
5) Call buildTree for elements after inIndex and make the built tree as right subtree of tNode.
6) return tNode.
Thanks to Rohini and Tushar for suggesting the code.
C++
#include <bits/stdc++.h>
using namespace std;
class node
{
public :
char data;
node* left;
node* right;
};
int search( char arr[], int strt, int end, char value);
node* newNode( char data);
node* buildTree( char in[], char pre[], int inStrt, int inEnd)
{
static int preIndex = 0;
if (inStrt > inEnd)
return NULL;
node* tNode = newNode(pre[preIndex++]);
if (inStrt == inEnd)
return tNode;
int inIndex = search(in, inStrt, inEnd, tNode->data);
tNode->left = buildTree(in, pre, inStrt, inIndex - 1);
tNode->right = buildTree(in, pre, inIndex + 1, inEnd);
return tNode;
}
int search( char arr[], int strt, int end, char value)
{
int i;
for (i = strt; i <= end; i++)
{
if (arr[i] == value)
return i;
}
}
node* newNode( char data)
{
node* Node = new node();
Node->data = data;
Node->left = NULL;
Node->right = NULL;
return (Node);
}
void printInorder(node* node)
{
if (node == NULL)
return ;
printInorder(node->left);
cout<<node->data<< " " ;
printInorder(node->right);
}
int main()
{
char in[] = { 'D' , 'B' , 'E' , 'A' , 'F' , 'C' };
char pre[] = { 'A' , 'B' , 'D' , 'E' , 'C' , 'F' };
int len = sizeof (in) / sizeof (in[0]);
node* root = buildTree(in, pre, 0, len - 1);
cout << "Inorder traversal of the constructed tree is " ;
printInorder(root);
}
C
#include <stdio.h>
#include <stdlib.h>
struct node {
char data;
struct node* left;
struct node* right;
};
int search( char arr[], int strt, int end, char value);
struct node* newNode( char data);
struct node* buildTree( char in[], char pre[], int inStrt, int inEnd)
{
static int preIndex = 0;
if (inStrt > inEnd)
return NULL;
struct node* tNode = newNode(pre[preIndex++]);
if (inStrt == inEnd)
return tNode;
int inIndex = search(in, inStrt, inEnd, tNode->data);
tNode->left = buildTree(in, pre, inStrt, inIndex - 1);
tNode->right = buildTree(in, pre, inIndex + 1, inEnd);
return tNode;
}
int search( char arr[], int strt, int end, char value)
{
int i;
for (i = strt; i <= end; i++) {
if (arr[i] == value)
return i;
}
}
struct node* newNode( char data)
{
struct node* node = ( struct node*) malloc ( sizeof ( struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
return (node);
}
void printInorder( struct node* node)
{
if (node == NULL)
return ;
printInorder(node->left);
printf ( "%c " , node->data);
printInorder(node->right);
}
int main()
{
char in[] = { 'D' , 'B' , 'E' , 'A' , 'F' , 'C' };
char pre[] = { 'A' , 'B' , 'D' , 'E' , 'C' , 'F' };
int len = sizeof (in) / sizeof (in[0]);
struct node* root = buildTree(in, pre, 0, len - 1);
printf ( "Inorder traversal of the constructed tree is " );
printInorder(root);
getchar ();
}
Java
class Node {
char data;
Node left, right;
Node( char item)
{
data = item;
left = right = null ;
}
}
class BinaryTree {
Node root;
static int preIndex = 0 ;
Node buildTree( char in[], char pre[], int inStrt, int inEnd)
{
if (inStrt > inEnd)
return null ;
Node tNode = new Node(pre[preIndex++]);
if (inStrt == inEnd)
return tNode;
int inIndex = search(in, inStrt, inEnd, tNode.data);
tNode.left = buildTree(in, pre, inStrt, inIndex - 1 );
tNode.right = buildTree(in, pre, inIndex + 1 , inEnd);
return tNode;
}
int search( char arr[], int strt, int end, char value)
{
int i;
for (i = strt; i <= end; i++) {
if (arr[i] == value)
return i;
}
return i;
}
void printInorder(Node node)
{
if (node == null )
return ;
printInorder(node.left);
System.out.print(node.data + " " );
printInorder(node.right);
}
public static void main(String args[])
{
BinaryTree tree = new BinaryTree();
char in[] = new char [] { 'D' , 'B' , 'E' , 'A' , 'F' , 'C' };
char pre[] = new char [] { 'A' , 'B' , 'D' , 'E' , 'C' , 'F' };
int len = in.length;
Node root = tree.buildTree(in, pre, 0 , len - 1 );
System.out.println( "Inorder traversal of constructed tree is : " );
tree.printInorder(root);
}
}
Python
class Node:
def __init__( self , data):
self .data = data
self .left = None
self .right = None
def buildTree(inOrder, preOrder, inStrt, inEnd):
if (inStrt > inEnd):
return None
tNode = Node(preOrder[buildTree.preIndex])
buildTree.preIndex + = 1
if inStrt = = inEnd :
return tNode
inIndex = search(inOrder, inStrt, inEnd, tNode.data)
tNode.left = buildTree(inOrder, preOrder, inStrt, inIndex - 1 )
tNode.right = buildTree(inOrder, preOrder, inIndex + 1 , inEnd)
return tNode
def search(arr, start, end, value):
for i in range (start, end + 1 ):
if arr[i] = = value:
return i
def printInorder(node):
if node is None :
return
printInorder(node.left)
print node.data,
printInorder(node.right)
inOrder = [ 'D' , 'B' , 'E' , 'A' , 'F' , 'C' ]
preOrder = [ 'A' , 'B' , 'D' , 'E' , 'C' , 'F' ]
buildTree.preIndex = 0
root = buildTree(inOrder, preOrder, 0 , len (inOrder) - 1 )
print "Inorder traversal of the constructed tree is"
printInorder(root)
C#
using System;
public class Node {
public char data;
public Node left, right;
public Node( char item)
{
data = item;
left = right = null ;
}
}
class GFG {
public Node root;
public static int preIndex = 0;
public virtual Node buildTree( char [] arr, char [] pre,
int inStrt, int inEnd)
{
if (inStrt > inEnd) {
return null ;
}
Node tNode = new Node(pre[preIndex++]);
if (inStrt == inEnd) {
return tNode;
}
int inIndex = search(arr, inStrt,
inEnd, tNode.data);
tNode.left = buildTree(arr, pre, inStrt, inIndex - 1);
tNode.right = buildTree(arr, pre, inIndex + 1, inEnd);
return tNode;
}
public virtual int search( char [] arr, int strt,
int end, char value)
{
int i;
for (i = strt; i <= end; i++) {
if (arr[i] == value) {
return i;
}
}
return i;
}
public virtual void printInorder(Node node)
{
if (node == null ) {
return ;
}
printInorder(node.left);
Console.Write(node.data + " " );
printInorder(node.right);
}
public static void Main( string [] args)
{
GFG tree = new GFG();
char [] arr = new char [] { 'D' , 'B' , 'E' , 'A' , 'F' , 'C' };
char [] pre = new char [] { 'A' , 'B' , 'D' , 'E' , 'C' , 'F' };
int len = arr.Length;
Node root = tree.buildTree(arr, pre, 0, len - 1);
Console.WriteLine( "Inorder traversal of "
+ "constructed tree is : " );
tree.printInorder(root);
}
}
Output:
Inorder traversal of the constructed tree is D B E A F C
Time Complexity: O(n^2). Worst case occurs when tree is left skewed. Example Preorder and Inorder traversals for worst case are {A, B, C, D} and {D, C, B, A}.
Efficient Approach :
We can optimize the above solution using hashing (unordered_map in C++ or HashMap in Java). We store indexes of inorder traversal in a hash table. So that search can be done O(1) time.
C++
#include <bits/stdc++.h>
using namespace std;
struct Node {
char data;
struct Node* left;
struct Node* right;
};
struct Node* newNode( char data)
{
struct Node* node = new Node;
node->data = data;
node->left = node->right = NULL;
return (node);
}
struct Node* buildTree( char in[], char pre[], int inStrt,
int inEnd, unordered_map< char , int >& mp)
{
static int preIndex = 0;
if (inStrt > inEnd)
return NULL;
char curr = pre[preIndex++];
struct Node* tNode = newNode(curr);
if (inStrt == inEnd)
return tNode;
int inIndex = mp[curr];
tNode->left = buildTree(in, pre, inStrt, inIndex - 1, mp);
tNode->right = buildTree(in, pre, inIndex + 1, inEnd, mp);
return tNode;
}
struct Node* buldTreeWrap( char in[], char pre[], int len)
{
unordered_map< char , int > mp;
for ( int i = 0; i < len; i++)
mp[in[i]] = i;
return buildTree(in, pre, 0, len - 1, mp);
}
void printInorder( struct Node* node)
{
if (node == NULL)
return ;
printInorder(node->left);
printf ( "%c " , node->data);
printInorder(node->right);
}
int main()
{
char in[] = { 'D' , 'B' , 'E' , 'A' , 'F' , 'C' };
char pre[] = { 'A' , 'B' , 'D' , 'E' , 'C' , 'F' };
int len = sizeof (in) / sizeof (in[0]);
struct Node* root = buldTreeWrap(in, pre, len);
printf ( "Inorder traversal of the constructed tree is " );
printInorder(root);
}
Output:
Inorder traversal of the constructed tree is D B E A F C
Time Complexity : O(n)
Another approach :
Use the fact that InOrder traversal is Left-Root-Right and PreOrder traversal is Root-Left-Right. Also, first node in the PreOrder traversal is always the root node and the first node in the InOrder traversal is the leftmost node in the tree.
Maintain two data-structures : Stack (to store the path we visited while traversing PreOrder array) and Set (to maintain the node in which the next right subtree is expected).
1. Do below until you reach the leftmost node.
Keep creating the nodes from PreOrder traversal
If the stack's topmost element is not in the set, link the created node to the left child of stack's topmost element (if any), without popping the element.
Else link the created node to the right child of stack's topmost element. Remove the stack's topmost element from the set and the stack.
Push the node to a stack.
2. Keep popping the nodes from the stack until either the stack is empty, or the topmost element of stack compares to the current element of InOrder traversal. Once the loop is over, push the last node back into the stack and into the set.
3. Goto Step 1.
import java.util.*;
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode( int x) { val = x; }
}
class BinaryTree {
static Set<TreeNode> set = new HashSet<>();
static Stack<TreeNode> stack = new Stack<>();
public TreeNode buildTree( int [] preorder, int [] inorder)
{
TreeNode root = null ;
for ( int pre = 0 , in = 0 ; pre < preorder.length;) {
TreeNode node = null ;
do {
node = new TreeNode(preorder[pre]);
if (root == null ) {
root = node;
}
if (!stack.isEmpty()) {
if (set.contains(stack.peek())) {
set.remove(stack.peek());
stack.pop().right = node;
}
else {
stack.peek().left = node;
}
}
stack.push(node);
} while (preorder[pre++] != inorder[in] && pre < preorder.length);
node = null ;
while (!stack.isEmpty() && in < inorder.length &&
stack.peek().val == inorder[in]) {
node = stack.pop();
in++;
}
if (node != null ) {
set.add(node);
stack.push(node);
}
}
return root;
}
void printInorder(TreeNode node)
{
if (node == null )
return ;
printInorder(node.left);
System.out.print(node.val + " " );
printInorder(node.right);
}
public static void main(String args[])
{
BinaryTree tree = new BinaryTree();
int in[] = new int [] { 9 , 8 , 4 , 2 , 10 , 5 , 10 , 1 , 6 , 3 , 13 , 12 , 7 };
int pre[] = new int [] { 1 , 2 , 4 , 8 , 9 , 5 , 10 , 10 , 3 , 6 , 7 , 12 , 13 };
int len = in.length;
TreeNode root = tree.buildTree(pre, in);
tree.printInorder(root);
}
}
Output:
9 8 4 2 10 5 10 1 6 3 13 12 7
Thanks Hardik Agarwal for suggesting this approach.
Construct a Binary Tree from Postorder and Inorder
Please write comments if you find any bug in above codes/algorithms, or find other ways to solve the same problem.
0 0
You Might Also Like
Subscribe to Our Newsletter
hennessywhavillat81.blogspot.com
Source: https://tutorialspoint.dev/data-structure/binary-tree-data-structure/construct-tree-from-given-inorder-and-preorder-traversal