In this post, we are going to solve the Construct Binary Tree from Preorder and Inorder Traversal Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.
Problem
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 <= 3000inorder.length == preorder.length-3000 <= preorder[i], inorder[i] <= 3000preorder andÂinorder consist of unique values.- Each value ofÂ
inorder also appears inÂpreorder. preorder is guaranteed to be the preorder traversal of the tree.inorder is guaranteed to be the inorder traversal of the tree.
Now, let’s see the leetcode solution of Construct Binary Tree from Preorder and Inorder Traversal Leetcode Solution.
Construct Binary Tree from Preorder and Inorder Traversal Leetcode Solution in Python
class Solution:
def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
inToIndex = {num: i for i, num in enumerate(inorder)}
def build(preStart: int, preEnd: int, inStart: int, inEnd: int) -> Optional[TreeNode]:
if preStart > preEnd:
return None
rootVal = preorder[preStart]
rootInIndex = inToIndex[rootVal]
leftSize = rootInIndex - inStart
root = TreeNode(rootVal)
root.left = build(preStart + 1, preStart + leftSize,
inStart, rootInIndex - 1)
root.right = build(preStart + leftSize + 1,
preEnd, rootInIndex + 1, inEnd)
return root
return build(0, len(preorder) - 1, 0, len(inorder) - 1)Construct Binary Tree from Preorder and Inorder Traversal Leetcode Solution in CPP
class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
unordered_map<int, int> inToIndex;
for (int i = 0; i < inorder.size(); ++i)
inToIndex[inorder[i]] = i;
return build(preorder, 0, preorder.size() - 1, inorder, 0,
inorder.size() - 1, inToIndex);
}
private:
TreeNode* build(const vector<int>& preorder, int preStart, int preEnd,
const vector<int>& inorder, int inStart, int inEnd,
const unordered_map<int, int>& inToIndex) {
if (preStart > preEnd)
return nullptr;
const int rootVal = preorder[preStart];
const int rootInIndex = inToIndex.at(rootVal);
const int leftSize = rootInIndex - inStart;
TreeNode* root = new TreeNode(rootVal);
root->left = build(preorder, preStart + 1, preStart + leftSize, inorder,
inStart, rootInIndex - 1, inToIndex);
root->right = build(preorder, preStart + leftSize + 1, preEnd, inorder,
rootInIndex + 1, inEnd, inToIndex);
return root;
}
};Construct Binary Tree from Preorder and Inorder Traversal Leetcode Solution in Java
class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
Map<Integer, Integer> inToIndex = new HashMap<>();
for (int i = 0; i < inorder.length; ++i)
inToIndex.put(inorder[i], i);
return build(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1, inToIndex);
}
private TreeNode build(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart,
int inEnd, Map<Integer, Integer> inToIndex) {
if (preStart > preEnd)
return null;
final int rootVal = preorder[preStart];
final int rootInIndex = inToIndex.get(rootVal);
final int leftSize = rootInIndex - inStart;
TreeNode root = new TreeNode(rootVal);
root.left = build(preorder, preStart + 1, preStart + leftSize, inorder, inStart,
rootInIndex - 1, inToIndex);
root.right = build(preorder, preStart + leftSize + 1, preEnd, inorder, rootInIndex + 1, inEnd,
inToIndex);
return root;
}
}Note: This problem Construct Binary Tree from Preorder and Inorder Traversal is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.
