In this post, you will know how to solve the Construct Binary Tree from Inorder and Postorder 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 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 <= 3000postorder.length == inorder.length-3000 <= inorder[i], postorder[i] <= 3000inorderandpostorderconsist of unique values.- Each value of
postorderalso appears ininorder. inorderis guaranteed to be the inorder traversal of the tree.postorderis guaranteed to be the postorder traversal of the tree.
Construct Binary Tree from Inorder and Postorder Traversal Leetcode Solutions 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)Construct Binary Tree from Inorder and Postorder Traversal Leetcode Solutions in CPP
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 Leetcode Solutions 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;
}
}Note: This problem Construct Binary Tree from Inorder and Postorder Traversal is generated by Leetcode but the solution is provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.
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