105. Construct Binary Tree from Preorder and Inorder Traversal LeetCode Solution

In this guide, you will get 105. Construct Binary Tree from Preorder and Inorder Traversal LeetCode Solution with the best time and space complexity. The solution to Construct Binary Tree from Preorder and Inorder Traversal problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Construct Binary Tree from Preorder and Inorder Traversal solution in C++
Construct Binary Tree from Preorder and Inorder Traversal solution in Java
Construct Binary Tree from Preorder and Inorder Traversal solution in Python
Additional Resources

105. Construct Binary Tree from Preorder and Inorder Traversal LeetCode Solution image

Problem Statement of Construct Binary Tree from Preorder and Inorder Traversal

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.

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

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

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;
  }
};
/* code provided by PROGIEZ */

105. 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;
  }
}
// code provided by PROGIEZ

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

class Solution:
  def buildTree(
      self,
      preorder: list[int],
      inorder: list[int],
  ) -> TreeNode | None:
    inToIndex = {num: i for i, num in enumerate(inorder)}

    def build(
        preStart: int,
        preEnd: int,
        inStart: int,
        inEnd: int,
    ) -> TreeNode | None:
      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)
 # code by PROGIEZ