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# Construct Binary Tree from Preorder and Inorder Traversal LeetCode Solution

## Problem – Construct Binary Tree from Preorder and Inorder Traversal LeetCode Solution

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.

### Construct Binary Tree from Preorder and Inorder Traversal LeetCode Solution in Java

``````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;
}
``````

### Construct Binary Tree from Preorder and Inorder Traversal LeetCode Solution in Python

``````def buildTree(self, preorder, inorder):
if inorder:
ind = inorder.index(preorder.pop(0))
root = TreeNode(inorder[ind])
root.left = self.buildTree(preorder, inorder[0:ind])
root.right = self.buildTree(preorder, inorder[ind+1:])
return root
``````

### Construct Binary Tree from Preorder and Inorder Traversal LeetCode Solution in C++

``````class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
int rootIdx = 0;
return build(preorder, inorder, rootIdx, 0, inorder.size()-1);
}

TreeNode* build(vector<int>& preorder, vector<int>& inorder, int& rootIdx, int left, int right) {
if (left > right) return NULL;
int pivot = left;  // find the root from inorder
while(inorder[pivot] != preorder[rootIdx]) pivot++;

rootIdx++;
TreeNode* newNode = new TreeNode(inorder[pivot]);
newNode->left = build(preorder, inorder, rootIdx, left, pivot-1);
newNode->right = build(preorder, inorder, rootIdx, pivot+1, right);
return newNode;
}
};
``````
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