Construct Binary Tree from Inorder and Postorder Traversal LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Link for the Problem – Construct 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:

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<int>& inorder, vector<int>& postorder) {
    unordered_map<int, int> 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<int>& inorder, int inStart, int inEnd,
                  const vector<int>& postorder, int postStart, int postEnd,
                  const unordered_map<int, int>& 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<Integer, Integer> 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<Integer, Integer> 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)