# Construct Binary Tree from Preorder and Inorder Traversal Problem

## Description

LeetCode Problem 105.

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:

``````1
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Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]
``````

Example 2:

``````1
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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.

## Sample C++ Code

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class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
if (preorder.size() == 0)
return NULL;
TreeNode* root = new TreeNode(preorder[0]);

vector<int> left_pre, right_pre, left_in, right_in;
bool isleft = true;
for (int i = 0; i < inorder.size(); i ++) {
if (inorder[i] == preorder[0]) {
isleft = false;
continue;
}
if (isleft) {
left_pre.push_back(preorder[i+1]);
left_in.push_back(inorder[i]);
} else {
right_pre.push_back(preorder[i]);
right_in.push_back(inorder[i]);
}
}

root->left = buildTree(left_pre, left_in);
root->right = buildTree(right_pre, right_in);

return root;
}
};
``````