105. Construct Binary Tree from Preorder and Inorder Traversal

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

 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private static int idx = 0;
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (null == preorder || preorder.length < 1) {
            return null;
        }
        
        int[] idx = {0};
        
        return buildTree(preorder, idx, inorder, 0, inorder.length - 1);
    }
    
    private TreeNode buildTree(int[] preorder, int[] idx, int[] inorder, int s, int e) {
        int rIdx = s;
        for (int i = s; i <= e; i++) {
            if (preorder[idx[0]] == inorder[i]) {
                rIdx = i;
                break;
            }
        }

        TreeNode root = new TreeNode(preorder[idx[0]]);
        if (rIdx - s >= 1) {
            idx[0] += 1;
            root.left = buildTree(preorder, idx, inorder, s, rIdx - 1);
        }
        
        if (e - rIdx >= 1) {
            idx[0] += 1;
            root.right = buildTree(preorder, idx, inorder, rIdx + 1, e);
        }
        
        return root;   
    }
}