Leetcode 105: Construct Binary Tree from Preorder and Inorder Traversal

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

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

 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode BuildTree(int[] preorder, int[] inorder) {
        if (preorder.Length == 0) return null;
        
        return DFS(preorder, 0, preorder.Length - 1, inorder, 0, inorder.Length - 1);
    }
    
    private TreeNode DFS(int[] preorder, int pStart, int pEnd, int[] inorder, int iStart, int iEnd)
    {
        if (pStart > pEnd) return null;
        if (pStart == pEnd) return new TreeNode(preorder[pStart]);
        
        var root = new TreeNode(preorder[pStart]);
        
        int i = iStart;
        for (; i <= iEnd; i++)
        {
            if (inorder[i] == preorder[pStart]) break;
        }
        
        root.left = DFS(preorder, pStart + 1, pStart + i - iStart, inorder, iStart, i - 1);
        root.right = DFS(preorder, pStart + i - iStart + 1, pEnd, inorder, i + 1, iEnd);
        
        return root;
    }
}