Construct Binary Tree from Preorder and Inorder Traversal

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

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

Example:

Given in-order [1,2,3] and pre-order [2,1,3], return a tree:

  2
 / \
1   3

Solution:

As can be seen on above example, we could repeat following steps to obtain a binary tree:

1. due to the definition of pre-order, we have the root value is preorder[0] (2)
2. use root value to find its index in in-order (1), then we know the left subtree is from 0 to 1 - 1 and right subtree is 1 + 1 to 2 which are inorder[0] (1) and inorder[3] respectively
3. based on the length of left and right subtree we can split preorder array into two subtrees

Code:

public class Solution {

    public TreeNode buildTree(int[] inorder, int[] preorder) {
        return buildTree(preorder, 0, preorder.length - 1, inorder, 
                        0, inorder.length - 1);
    }

    private TreeNode buildTree(int[] preorder, int pstart, int pend,
                                int[] inorder, int istart, int iend) {
        if (pstart > pend || istart > iend) {
            return null;
        }
        int idx = findIndex(preorder, pstart, pend, inorder[istart]);
        TreeNode node = new TreeNode(preorder[idx]);
        node.left = buildTree(preorder, pstart, idx - 1, inorder,
                                istart + 1, idx - pstart + istart);
        node.right = buildTree(preorder, idx + 1, pend, inorder,
                                idx - pstart + istart + 1, iend);
        return node;
    }

    private int findIndex(int[] preorder, int pstart, int pend, int num) {
        int i = 0;
        for (i = pstart; i <= pend; i++) {
            if (preorder[i] == num) {
                break;
            }
        }
        return i;
    }
}