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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Construct Binary Tree from Inorder and Postorder Traversal in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemConstruct Binary Tree from Inorder and Postorder Traversal– LeetCode Problem

Construct Binary Tree from Inorder and Postorder Traversal– LeetCode Problem

Problem:

Given two integer arrays inorder and postorder where inorder is the inorder traversal of a binary tree and postorder is the postorder traversal of the same tree, construct and return the binary tree.

Example 1:

tree
Input: inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
Output: [3,9,20,null,null,15,7]

Example 2:

Input: inorder = [-1], postorder = [-1]
Output: [-1]

Constraints:

  • 1 <= inorder.length <= 3000
  • postorder.length == inorder.length
  • -3000 <= inorder[i], postorder[i] <= 3000
  • inorder and postorder consist of unique values.
  • Each value of postorder also appears in inorder.
  • inorder is guaranteed to be the inorder traversal of the tree.
  • postorder is guaranteed to be the postorder traversal of the tree.
Construct Binary Tree from Inorder and Postorder Traversal– LeetCode Solutions
Construct Binary Tree from Inorder and Postorder Traversal Solution in C++:
class Solution {
 public:
  TreeNode* buildTree(vector& inorder, vector& postorder) {
    unordered_map inToIndex;

    for (int i = 0; i < inorder.size(); ++i)
      inToIndex[inorder[i]] = i;

    return build(inorder, 0, inorder.size() - 1, postorder, 0,
                 postorder.size() - 1, inToIndex);
  }

 private:
  TreeNode* build(const vector& inorder, int inStart, int inEnd,
                  const vector& postorder, int postStart, int postEnd,
                  const unordered_map& inToIndex) {
    if (inStart > inEnd)
      return nullptr;

    const int rootVal = postorder[postEnd];
    const int rootInIndex = inToIndex.at(rootVal);
    const int leftSize = rootInIndex - inStart;

    TreeNode* root = new TreeNode(rootVal);
    root->left = build(inorder, inStart, rootInIndex - 1, postorder, postStart,
                       postStart + leftSize - 1, inToIndex);
    root->right = build(inorder, rootInIndex + 1, inEnd, postorder,
                        postStart + leftSize, postEnd - 1, inToIndex);
    return root;
  }
};
Construct Binary Tree from Inorder and Postorder Traversal Solution in Java:
class Solution {
  public TreeNode buildTree(int[] inorder, int[] postorder) {
    Map inToIndex = new HashMap<>();

    for (int i = 0; i < inorder.length; ++i)
      inToIndex.put(inorder[i], i);

    return build(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1, inToIndex);
  }

  TreeNode build(int[] inorder, int inStart, int inEnd, int[] postorder, int postStart, int postEnd,
                 Map inToIndex) {
    if (inStart > inEnd)
      return null;

    final int rootVal = postorder[postEnd];
    final int rootInIndex = inToIndex.get(rootVal);
    final int leftSize = rootInIndex - inStart;

    TreeNode root = new TreeNode(rootVal);
    root.left = build(inorder, inStart, rootInIndex - 1, postorder, postStart,
                      postStart + leftSize - 1, inToIndex);
    root.right = build(inorder, rootInIndex + 1, inEnd, postorder, postStart + leftSize,
                       postEnd - 1, inToIndex);
    return root;
  }
}
Construct Binary Tree from Inorder and Postorder Traversal Solution in Python:
class Solution:
  def buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
    inToIndex = {num: i for i, num in enumerate(inorder)}

    def build(inStart: int, inEnd: int, postStart: int, postEnd: int) -> Optional[TreeNode]:
      if inStart > inEnd:
        return None

      rootVal = postorder[postEnd]
      rootInIndex = inToIndex[rootVal]
      leftSize = rootInIndex - inStart

      root = TreeNode(rootVal)
      root.left = build(inStart, rootInIndex - 1,  postStart,
                        postStart + leftSize - 1)
      root.right = build(rootInIndex + 1, inEnd,  postStart + leftSize,
                         postEnd - 1)
      return root

    return build(0, len(inorder) - 1, 0, len(postorder) - 1)

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