Construct Binary Tree from Preorder and Inorder Traversal | Leetcode

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

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

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Solution: The first node in the preorder is always the root node. Find this same node in the inorder. All the nodes to the left belong to the left sub-tree. All the nodes to the right belong to the right sub-tree. In each recursion, always pick the next node from the preorder as the root node.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private final Map<Integer, Integer> inOrderIndexMap = new HashMap<>();
    private int preOrderIndex = 0;
    
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder.length == 0) return null;
        for (int i = 0; i < inorder.length; i++) {
            inOrderIndexMap.put(inorder[i], i);
        }
        return buildTree(preorder, inorder, 0, inorder.length - 1);
    }
    
    private TreeNode buildTree(int[] preorder, int[] inorder, int i0, int i1) {
        if (i0 > i1) return null;
        TreeNode root = new TreeNode(preorder[preOrderIndex++]);
        int rootIndex = inOrderIndexMap.get(root.val);
        root.left = buildTree(preorder, inorder, i0, rootIndex - 1);
        root.right = buildTree(preorder, inorder, rootIndex + 1, i1);
        return root;
    }
}