0105. Construct Binary Tree from Preorder and Inorder Traversal

https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal

Description

Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.

Example 1:

**Input:** preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
**Output:** [3,9,20,null,null,15,7]

Example 2:

**Input:** preorder = [-1], inorder = [-1]
**Output:** [-1]

Constraints:

• 1 <= preorder.length <= 3000

• inorder.length == preorder.length

• -3000 <= preorder[i], inorder[i] <= 3000

• preorder and inorder consist of unique values.

• Each value of inorder also appears in preorder.

• preorder is guaranteed to be the preorder traversal of the tree.

• inorder is guaranteed to be the inorder traversal of the tree.

ac

// Tree intuition, divide & conquer recursion
// find root index in in-order is the key: if (preorder[pstart] == inorder[i]) rootIdx = i;

/**
 * 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) {
        // pre-order 1st -> find mid int in-order
        // divide left / right parts, recursion

        // edge cases
        if (preorder.length == 0 || inorder.length == 0) return null;

        return helper(preorder, 0, preorder.length-1, inorder, 0, inorder.length-1);
    }

    private TreeNode helper(int[] preorder, int pstart, int pend, int[] inorder, int istart, int iend) {
        // exit
        if (pstart > pend || istart > iend) return null;
        if (pstart == pend || istart == iend) return new TreeNode(preorder[pstart]);

        // recursion
        // find root index
        int rootIdx = -1;
        for (int i = istart; i <= iend; i++) {
            if (preorder[pstart] == inorder[i]){
                rootIdx = i;
                break;
            }
        }
        int len = rootIdx - istart;

        // divdide
        TreeNode root = new TreeNode(inorder[rootIdx]);
        root.left = helper(preorder, pstart+1, pstart+len, inorder, istart, rootIdx-1);
        root.right = helper(preorder, pstart+len+1, pend, inorder, rootIdx+1, iend);

        return root;
    }
}