106. Construct Binary Tree from Inorder and Postorder Traversal
Read the full problem statement on LeetCode.
Difficulty: medium Acceptance: 66% Topics: Array, Hash Table, Divide and Conquer, Tree, Binary Tree
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Reference solution (spoiler · python)
# Time: O(n)
# Space: O(n)
class TreeNode(object):
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution(object):
# @param inorder, a list of integers
# @param postorder, a list of integers
# @return a tree node
def buildTree(self, inorder, postorder):
lookup = {}
for i, num in enumerate(inorder):
lookup[num] = i
return self.buildTreeRecu(lookup, postorder, inorder, len(postorder), 0, len(inorder))
def buildTreeRecu(self, lookup, postorder, inorder, post_end, in_start, in_end):
if in_start == in_end:
return None
node = TreeNode(postorder[post_end - 1])
i = lookup[postorder[post_end - 1]]
node.left = self.buildTreeRecu(lookup, postorder, inorder, post_end - 1 - (in_end - i - 1), in_start, i)
node.right = self.buildTreeRecu(lookup, postorder, inorder, post_end - 1, i + 1, in_end)
return node
Solution from kamyu104/LeetCode-Solutions · MIT
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