# Construct Binary Tree from Inorder and Postorder Traversal Leetcode Solution

In this post, we are going 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 <= 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.

Now, lets see the leetcode solution of Construct Binary Tree from Inorder and Postorder Traversal Leetcode Solution.

### Construct Binary Tree from Inorder and Postorder Traversal Leetcode 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)

### Construct Binary Tree from Inorder and Postorder Traversal Leetcode Solutionin 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 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;
}
}

Note: This problem Construct Binary Tree from Inorder and Postorder Traversal is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

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