# 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:**

![](https://assets.leetcode.com/uploads/2021/02/19/tree.jpg)

```
**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

```java
// 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;
    }
}
```


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