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# 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.

For example, given

```
preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]
```

Return the following binary tree:

```
    3
   / \
  9  20
    /  \
   15   7
```

## Solution

The basic idea is here:\
Say we have 2 arrays, PRE and IN.\
Preorder traversing implies that PRE\[0] is the root node.\
Then we can find this PRE\[0] in IN, say it's IN\[5].\
Now we know that IN\[5] is root, so we know that IN\[0] - IN\[4] is on the left side, IN\[6] to the end is on the right side.\
Recursively doing this on subarrays, we can build a tree out of it.

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

public TreeNode helper(int preStart, int inStart, int inEnd, int[] preorder, int[] inorder) {
    if (preStart > preorder.length - 1 || inStart > inEnd) {
        return null;
    }
    TreeNode root = new TreeNode(preorder[preStart]);
    int inIndex = 0; // Index of current root in inorder
    for (int i = inStart; i <= inEnd; i++) {
        if (inorder[i] == root.val) {
            inIndex = i;
        }
    }
    root.left = helper(preStart + 1, inStart, inIndex - 1, preorder, inorder);
    root.right = helper(preStart + inIndex - inStart + 1, inIndex + 1, inEnd, preorder, inorder);
    return root;
}
```


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