LeetCode Top Interview 150

106. Construct Binary Tree from Inorder and Postorder Traversal

Medium

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.

Solution

from typing import List, Optional

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
        in_index = [len(inorder) - 1]
        post_index = [len(postorder) - 1]
        return self._helper(inorder, postorder, in_index, post_index, float('inf'))

    def _helper(self, inorder: List[int], postorder: List[int], index: List[int], p_index: List[int], target: int) -> Optional[TreeNode]:
        if index[0] < 0 or (index[0] < len(inorder) and inorder[index[0]] == target):
            return None
        root = TreeNode(postorder[p_index[0]])
        p_index[0] -= 1
        root.right = self._helper(inorder, postorder, index, p_index, root.val)
        index[0] -= 1
        root.left = self._helper(inorder, postorder, index, p_index, target)
        return root

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