LeetCode Java Solutions

105. Construct Binary Tree from Preorder and Inorder Traversal

Medium

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

Note:
You may assume that duplicates do not exist in the tree.

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

3
   / \
  9  20
    /  \
   15   7

Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        return helper(0, 0, inorder.length - 1, preorder, inorder);
    }
    private TreeNode helper(int preStart, int inStart, int inEnd, int[] preOrder, int[] inOrder) {
       //Boundary Conditions.
       if(preStart > preOrder.length-1) {
          return null;
       }
       if(inStart > inEnd) {
          return null;
       }
       TreeNode root = new TreeNode(preOrder[preStart]);
       int inIndex = 0;
       for(int i=inStart; i<= inEnd; i++) {
          if(root.val == inOrder[i]) {
             inIndex = i;
          }
       }
       root.left = helper(preStart+1, inStart, inIndex-1, preOrder, inOrder);
       root.right = helper(preStart + inIndex-inStart +1, inIndex+1, inEnd, preOrder,inOrder);
       return root;
   }
}