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

**Input:** Inorder traversal and Depth-First-Search.

**Approach:**

**int**[] inOrder = { 8, 4, 2, 5, 1, 6, 3, 7, 9 };

**int**[] DFS = { 1, 2, 4, 8, 5, 3, 6, 7, 9 };

- First element in DFS
*[]*will be the*root*of the tree, here its 1. - Now the search element 1 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 InorderDFStoTree { | |

int indexDFS = 0; | |

public Node makeBTree(int[] inorder, int[] DFS, int iStart, int iEnd) { | |

if (iStart > iEnd) { | |

return null; | |

} | |

int rootVal = DFS[indexDFS]; | |

Node root = new Node(rootVal); | |

indexDFS++; | |

if (iStart == iEnd) { | |

return root; | |

} | |

int index = findIndex(inorder, rootVal, iStart, iEnd); | |

root.left = makeBTree(inorder, DFS, iStart, index – 1); | |

root.right = makeBTree(inorder, DFS, index + 1, iEnd); | |

return root; | |

} | |

public int findIndex(int[] inorder, int value, int iStart, int iEnd) { | |

int x = –1; | |

for (int i = iStart; i <= iEnd; i++) { | |

if (value == inorder[i]) { | |

x = i; | |

} | |

} | |

return x; | |

} | |

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[]) { | |

int[] inOrder = { 8, 4, 2, 5, 1, 6, 3, 7, 9 }; | |

int[] DFS = { 1, 2, 4, 8, 5, 3, 6, 7, 9 }; | |

InorderDFStoTree i = new InorderDFStoTree(); | |

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

System.out.println("inorder traversal of 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:**

inorder traversal of constructed tree : 8 4 2 5 1 6 3 7 9