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원문:https://www.geeksforgeeks.org/construct-tree-inorder-level-order-traversals/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

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.

  // Java program to construct a tree from level order and 
    // and inorder traversal 
   
// A binary tree node 
class 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 Output:

Inorder traversal of the constructed tree is
4 8 10 12 14 20 22

https://www.geeksforgeeks.org/construct-tree-inorder-level-order-traversals/