Construct Binary Tree from preorder and inorder traversal

Given Preorder and inorder traversal of a tree, construct the binary tree.

Note:
Assume that duplicates does not exist in the tree.

Subscribe to see which companies asked this question

Hide TagsTree Array Depth-first SearchHide Similar Problems(M) Construct Binary Tree from Inorder and Postorder traversalOriginal version, easier to understand, and doing more copying work.

/*** Definition for a binary tree node. * public class TreeNode {* int val; * TreeNode left; * TreeNode rig Ht * TreeNode (int x) {val = x;} }*/ Public classSolution { PublicTreeNode Buildtree (int[] Preorder,int[] inorder) {        if(Preorder.length = = 0)            return NULL; intRootvalue = preorder[0]; TreeNode Root=NewTreeNode (Rootvalue); if(Preorder.length = = 1)            returnRoot; intInorderrootindex = 0;  for(inti = 0; i<inorder.length; ++i) {if(Inorder[i] = =rootvalue) {Inorderrootindex=i;  Break; }        }                int[] Preorderleftbranch = Arrays.copyofrange (preorder, 1, inorderrootindex+1); int[] Preorderrightbranch = Arrays.copyofrange (Preorder, inorderrootindex+1, preorder.length); int[] Inorderleftbranch = Arrays.copyofrange (inorder, 0, Inorderrootindex); int[] Inorderrightbranch = Arrays.copyofrange (inorder, inorderrootindex+1, inorder.length); Root.left=Buildtree (Preorderleftbranch, inorderleftbranch); Root.right=Buildtree (Preorderrightbranch, inorderrightbranch); returnRoot; }}

Improved version. Avoid array copying and improved root index look up.

/*** Definition for Binary tree * public class TreeNode {* Int. val; * TreeNode left; * TreeNode right; * TreeNode (int x) {val = x;} }*/ Public classSolution { PublicTreeNode Buildtree (int[] Preorder,int[] inorder) {Map<integer, integer> inmap =NewHashmap<integer, integer>();  for(inti = 0; i < inorder.length; i++) {inmap.put (inorder[i], i); } TreeNode Root= Buildtree (preorder, 0, preorder.length-1, inorder, 0, Inorder.length-1, Inmap); returnRoot; }         PublicTreeNode Buildtree (int[] Preorder,intPrestart,intPreend,int[] inorder,intInstart,intInend, Map<integer, integer>Inmap) {        if(Prestart > Preend | | instart > Inend)return NULL; TreeNode Root=NewTreeNode (Preorder[prestart]); intInroot =Inmap.get (Root.val); intNumsleft = Inroot-Instart; Root.left= Buildtree (Preorder, Prestart + 1, Prestart + numsleft, inorder, Instart, inRoot-1, Inmap); Root.right= Buildtree (preorder, Prestart + numsleft + 1, preend, inorder, Inroot + 1, Inend, Inmap); returnRoot; }}

Construct Binary Tree from preorder and inorder traversal