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

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.

Construct Binary Tree from Preorder and Inorder Traversal Solutions

✅Time: O(n)
✅Space: O(n)

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;
  }
};

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;
  }
}

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)

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