Construct a tree from Inorder and Level order traversals | Set 1 - GeeksforGeeks

21 Feb, 2022
Given inorder and level-order traversals of a Binary Tree, construct the Binary Tree. Following is an example to illustrate the problem.
Input: Two arrays that represent Inorder
       and level order traversals of a 
       Binary Tree
in[]    = {4, 8, 10, 12, 14, 20, 22};
level[] = {20, 8, 22, 4, 12, 10, 14};

Output: Construct the tree represented 
        by the two arrays.
        For the above two arrays, the 
        constructed tree is shown in 
        the diagram on right side
The following post can be considered as a prerequisite for this. Construct Tree from given Inorder and Preorder traversals 
Let us consider the above example.in[] = {4, 8, 10, 12, 14, 20, 22}; level[] = {20, 8, 22, 4, 12, 10, 14};In a Levelorder sequence, the first element is the root of the tree. So we know ’20’ is root for given sequences. By searching ’20’ in Inorder sequence, we can find out all elements on left side of ‘20’ are in left subtree and elements on right are in right subtree. So we know below structure now. 
             20
           /    \
          /      \ 
 {4,8,10,12,14}  {22}
Let us call {4,8,10,12,14} as left subarray in Inorder traversal and {22} as right subarray in Inorder traversal. In level order traversal, keys of left and right subtrees are not consecutive. So we extract all nodes from level order traversal which are in left subarray of Inorder traversal. To construct the left subtree of root, we recur for the extracted elements from level order traversal and left subarray of inorder traversal. In the above example, we recur for following two arrays. 
// Recur for following arrays to construct the left subtree
In[]    = {4, 8, 10, 12, 14}
level[] = {8, 4, 12, 10, 14}
Similarly, we recur for following two arrays and construct the right subtree.
// Recur for following arrays to construct the right subtree
In[]    = {22}
level[] = {22}
Following is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
/* program to construct tree using inorder and levelorder * traversals */#include <bits/stdc++.h>using namespace std; /* A binary tree node */struct Node {    int key;    struct Node *left, *right;}; /* Function to find index of value in arr[start...end] */int search(int arr[], int strt, int end, int value){    for (int i = strt; i <= end; i++)        if (arr[i] == value)            return i;    return -1;} // n is size of level[], m is size of in[] and m < n. This// function extracts keys from level[] which are present in// in[].  The order of extracted keys must be maintainedint* extrackKeys(int in[], int level[], int m, int n){    int *newlevel = new int[m], j = 0;    for (int i = 0; i < n; i++)        if (search(in, 0, m - 1, level[i]) != -1)            newlevel[j] = level[i], j++;    return newlevel;} /* function that allocates a new node with the given key  */Node* newNode(int key){    Node* node = new Node;    node->key = key;    node->left = node->right = NULL;    return (node);} /* Recursive function to construct binary tree of size n   from Inorder traversal in[] and Level Order traversal   level[]. inStrt and inEnd are start and end indexes of   array in[] Initial values of inStrt and inEnd should be 0   and n -1. The function doesn't do any error checking for   cases where inorder and levelorder do not form a tree */Node* buildTree(int in[], int level[], int inStrt,                int inEnd, int n){     // If start index is more than the end index    if (inStrt > inEnd)        return NULL;     /* The first node in level order traversal is root */    Node* root = newNode(level[0]);     /* If this node has no children then return */    if (inStrt == inEnd)        return root;     /* Else find the index of this node in Inorder traversal     */    int inIndex = search(in, inStrt, inEnd, root->key);     // Extract left subtree keys from level order traversal    int* llevel = extrackKeys(in, level, inIndex, n);     // Extract right subtree keys from level order traversal    int* rlevel        = extrackKeys(in + inIndex + 1, level, n - 1, n);     /* construct left and right subtrees */    root->left = buildTree(in, llevel, inStrt, inIndex - 1,                           inIndex - inStrt);    root->right = buildTree(in, rlevel, inIndex + 1, inEnd,                            inEnd - inIndex);     // Free memory to avoid memory leak    delete[] llevel;    delete[] rlevel;     return root;} /* utility function to print inorder traversal of binary * tree */void printInorder(Node* node){    if (node == NULL)        return;    printInorder(node->left);    cout << node->key << " ";    printInorder(node->right);} /* Driver program to test above functions */int main(){    int in[] = { 4, 8, 10, 12, 14, 20, 22 };    int level[] = { 20, 8, 22, 4, 12, 10, 14 };    int n = sizeof(in) / sizeof(in[0]);    Node* root = buildTree(in, level, 0, n - 1, n);     /* Let us test the built tree by printing Inorder     * traversal */    cout << "Inorder traversal of the constructed tree is "            "\n";    printInorder(root);     return 0;}
// Java program to construct a tree from level order and// and inorder traversal // A binary tree nodeclass Node {    int data;    Node left, right;     Node(int item)    {        data = item;        left = right = null;    }     public void setLeft(Node left) { this.left = left; }     public void setRight(Node right) { this.right = right; }} class Tree {    Node root;     Node buildTree(int in[], int level[])    {        Node startnode = null;        return constructTree(startnode, level, in, 0,                             in.length - 1);    }     Node constructTree(Node startNode, int[] levelOrder,                       int[] inOrder, int inStart,                       int inEnd)    {         // if start index is more than end index        if (inStart > inEnd)            return null;         if (inStart == inEnd)            return new Node(inOrder[inStart]);         boolean found = false;        int index = 0;         // it represents the index in inOrder array of        // element that appear first in levelOrder array.        for (int i = 0; i < levelOrder.length - 1; i++) {            int data = levelOrder[i];            for (int j = inStart; j < inEnd; j++) {                if (data == inOrder[j]) {                    startNode = new Node(data);                    index = j;                    found = true;                    break;                }            }            if (found == true)                break;        }         // elements present before index are part of left        // child of startNode. elements present after index        // are part of right child of startNode.        startNode.setLeft(            constructTree(startNode, levelOrder, inOrder,                          inStart, index - 1));        startNode.setRight(            constructTree(startNode, levelOrder, inOrder,                          index + 1, inEnd));         return startNode;    }     /* Utility function to print inorder traversal of binary     * tree */    void printInorder(Node node)    {        if (node == null)            return;        printInorder(node.left);        System.out.print(node.data + " ");        printInorder(node.right);    }     // Driver program to test the above functions    public static void main(String args[])    {        Tree tree = new Tree();        int in[] = new int[] { 4, 8, 10, 12, 14, 20, 22 };        int level[]            = new int[] { 20, 8, 22, 4, 12, 10, 14 };        int n = in.length;        Node node = tree.buildTree(in, level);         /* Let us test the built tree by printing Inorder         * traversal */        System.out.print(            "Inorder traversal of the constructed tree is ");        tree.printInorder(node);    }} // This code has been contributed by Mayank Jaiswal
# Python program to construct tree using# inorder and level order traversals # A binary tree node  class Node:     # Constructor to create a new node    def __init__(self, key):        self.data = key        self.left = None        self.right = None  """Recursive function to construct binary tree of size n fromInorder traversal ino[] and Level Order traversal level[].The function doesn't do any error checking for caseswhere inorder and levelorder do not form a tree """  def buildTree(level, ino):     # If ino array is not empty    if ino:         # Check if that element exist in level order        for i in range(0, len(level)):             if level[i] in ino:                 # Create a new node with                # the matched element                node = Node(level[i])                 # Get the index of the matched element                # in level order array                io_index = ino.index(level[i])                break         # Construct left and right subtree        node.left = buildTree(level, ino[0:io_index])        node.right = buildTree(level, ino[io_index + 1:len(ino)])        return node     else:        return None  def printInorder(node):    if node is None:        return     # first recur on left child    printInorder(node.left)     # then print the data of node    print(node.data, end=" ")     # now recur on right child    printInorder(node.right) # Driver code  levelorder = [20, 8, 22, 4, 12, 10, 14]inorder = [4, 8, 10, 12, 14, 20, 22] ino_len = len(inorder)root = buildTree(levelorder, inorder) # Let us test the build tree by# printing Inorder traversalprint("Inorder traversal of the constructed tree is")printInorder(root) # This code is contributed by 'Vaibhav Kumar'
// C# program to construct a tree from// level order and and inorder traversalusing System; // A binary tree nodepublic class Node {    public int data;    public Node left, right;     public Node(int item)    {        data = item;        left = right = null;    }     public virtual Node Left    {        set { this.left = value; }    }     public virtual Node Right    {        set { this.right = value; }    }} class GFG {    public Node root;     public virtual Node buildTree(int[] arr, int[] level)    {        Node startnode = null;        return constructTree(startnode, level, arr, 0,                             arr.Length - 1);    }     public virtual Node    constructTree(Node startNode, int[] levelOrder,                  int[] inOrder, int inStart, int inEnd)    {         // if start index is more than end index        if (inStart > inEnd) {            return null;        }         if (inStart == inEnd) {            return new Node(inOrder[inStart]);        }         bool found = false;        int index = 0;         // it represents the index in inOrder        // array of element that appear first        // in levelOrder array.        for (int i = 0; i < levelOrder.Length - 1; i++) {            int data = levelOrder[i];            for (int j = inStart; j < inEnd; j++) {                if (data == inOrder[j]) {                    startNode = new Node(data);                    index = j;                    found = true;                    break;                }            }            if (found == true) {                break;            }        }         // elements present before index are        // part of left child of startNode.        // elements present after index are        // part of right child of startNode.        startNode.Left            = constructTree(startNode, levelOrder, inOrder,                            inStart, index - 1);        startNode.Right            = constructTree(startNode, levelOrder, inOrder,                            index + 1, inEnd);         return startNode;    }     /* Utility function to print inorder       traversal of binary tree */    public virtual void printInorder(Node node)    {        if (node == null) {            return;        }        printInorder(node.left);        Console.Write(node.data + " ");        printInorder(node.right);    }     // Driver Code    public static void Main(string[] args)    {        GFG tree = new GFG();        int[] arr = new int[] { 4, 8, 10, 12, 14, 20, 22 };        int[] level            = new int[] { 20, 8, 22, 4, 12, 10, 14 };        int n = arr.Length;        Node node = tree.buildTree(arr, level);         /* Let us test the built tree by        printing Inorder traversal */        Console.Write("Inorder traversal of the "                      + "constructed tree is "                      + "\n");        tree.printInorder(node);    }} // This code is contributed by Shrikant13
<script> // inorder and level order traversals// A binary tree nodeclass Node{     // Constructor to create a new node    constructor(key){        this.data = key        this.left = null        this.right = null    } } // Recursive function to construct binary tree of size n from// Inorder traversal ino[] and Level Order traversal level[].// The function doesn't do any error checking for cases// where inorder and levelorder do not form a treefunction buildTree(level, ino){     // If ino array is not empty    if(ino.length > 0)    {             // Check if that element exist in level order        let io_index;        let node;        for (let i = 0; i < level.length; i++)        {             if(ino.indexOf(level[i]) !== -1)            {                 // Create a new node with                // the matched element                node = new Node(level[i])                 // Get the index of the matched element                // in level order array                io_index = ino.indexOf(level[i]);                break             }                     }                 // Construct left and right subtree        node.left = buildTree(level, ino.slice(0,io_index))        node.right = buildTree(level, ino.slice(io_index+1,ino.length))        return node    }     else return null; } function printInorder(node){    if(node === null)        return     // first recur on left child    printInorder(node.left)     // then print the data of node    document.write(node.data," ")     // now recur on right child    printInorder(node.right)} // Driver codelet levelorder = [20, 8, 22, 4, 12, 10, 14];let inorder = [4, 8, 10, 12, 14, 20, 22]; root = buildTree(levelorder, inorder); // Let us test the build tree by// printing Inorder traversaldocument.write("Inorder traversal of the constructed tree is","</br>");printInorder(root); // This code is contributed by shinjanpatra </script>
Output: 
Inorder traversal of the constructed tree is
4 8 10 12 14 20 22
An upper bound on time complexity of above method is O(n3). In the main recursive function, extractNodes() is called which takes O(n2) time.The code can be optimized in many ways and there may be better solutions. Construct a tree from Inorder and Level order traversals | Set 2This article is contributed by Abhay Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above 
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