**Objective:** **– **Given a inorder and preorder traversal, construct a binary tree from that.

**Input:** Inorder and preorder traversals

**Similar Problem: Construct a binary tree from given Inorder and Postorder Traversal**

**Approach:**

**int** [] inOrder = {2,5,6,10,12,14,15};

**int** [] preOrder = {10,5,2,6,14,12,15};

- First element in
*preorder[]*will be the*root*of the tree, here its 10. - Now the search element 10 in
*inorder[]*, say you find it at position*i*, once you find it, make note of elements which are left to*i*(this will construct the leftsubtree) and elements which are right to*i*( this will construct the rightSubtree). - See this step above and recursively construct left subtree and link it root.left and recursively construct right subtree and link it root.right.

- See the picture and code.

**Complete Code:**

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public class MakeTreefromInorderAndPreorder { | |

public static int pIndex=0; | |

public Node makeBTree(int [] inOrder, int [] preOrder, int iStart, int iEnd ){ | |

if(iStart>iEnd){ | |

return null; | |

} | |

Node root = new Node(preOrder[pIndex]);pIndex++; | |

if(iStart==iEnd){ | |

return root; | |

} | |

int index = getInorderIndex(inOrder, iStart, iEnd, root.data); | |

root.left = makeBTree(inOrder,preOrder,iStart, index–1); | |

root.right = makeBTree(inOrder,preOrder,index+1, iEnd); | |

//} | |

return root; | |

} | |

public int getInorderIndex(int [] inOrder, int start, int end, int data){ | |

for(int i=start;i<=end;i++){ | |

if(inOrder[i]==data){ | |

return i; | |

} | |

} | |

return –1; | |

} | |

public void printINORDER(Node root){ | |

if(root!=null){ | |

printINORDER(root.left); | |

System.out.print(" " + root.data); | |

printINORDER(root.right); | |

} | |

} | |

public static void main (String[] args) throws java.lang.Exception | |

{ | |

int [] inOrder = {2,5,6,10,12,14,15}; | |

int [] preOrder = {10,5,2,6,14,12,15}; | |

MakeTreefromInorderAndPreorder i = new MakeTreefromInorderAndPreorder(); | |

Node x = i.makeBTree(inOrder, preOrder, 0, inOrder.length–1); | |

System.out.println("Constructed Tree : "); | |

i.printINORDER(x); | |

} | |

} | |

class Node{ | |

int data; | |

Node left; | |

Node right; | |

public Node (int data){ | |

this.data = data; | |

left = null; | |

right = null; | |

} | |

} |

**Output**:

Constructed Tree : 2 5 6 10 12 14 15

Broken method.

No solution for —

ADFGHKLPQRWZ(preorder)

GFHKDLAWRQPZ(inorder)

GKHFLDWRQZPA (Postorder)