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106. Construct Binary Tree From Inorder And Postorder Traversal

Array Hash Table Divide and Conquer Tree Binary Tree

Problem - 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.

Solutions

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# Definition for a binary tree node.
# 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]:
        def dfs(i: int, j: int, n: int) -> Optional[TreeNode]:
            if n <= 0:
                return None
            val = postorder[j + n - 1]
            k = hash_table[val]
            left = dfs(i, j, k - i)
            right = dfs(k + 1, j + k - i, n - k + i - 1)

            return TreeNode(val, left, right)

        hash_table = {val: i for i, val in enumerate(inorder)}
        return dfs(0, 0, len(inorder))

Submission Stats:

  • Runtime: 5 ms (59.36%)
  • Memory: 19.2 MB (94.94%)