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construct-binary-tree-from-preorder-and-inorder-traversal.py 1.63 KB
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Sanghee Kim 提交于 2019-01-15 06:07 +08:00 . add a solution
# 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 preorder, a list of integers
# @param inorder, a list of integers
# @return a tree node
def buildTree(self, preorder, inorder):
lookup = {}
for i, num in enumerate(inorder):
lookup[num] = i
return self.buildTreeRecu(lookup, preorder, inorder, 0, 0, len(inorder))
def buildTreeRecu(self, lookup, preorder, inorder, pre_start, in_start, in_end):
if in_start == in_end:
return None
node = TreeNode(preorder[pre_start])
i = lookup[preorder[pre_start]]
node.left = self.buildTreeRecu(lookup, preorder, inorder, pre_start + 1, in_start, i)
node.right = self.buildTreeRecu(lookup, preorder, inorder, pre_start + 1 + i - in_start, i + 1, in_end)
return node
# time: O(n)
# space: O(n)
class Solution2(object):
def buildTree(self, preorder, inorder):
"""
:type preorder: List[int]
:type inorder: List[int]
:rtype: TreeNode
"""
preorder_iterator = iter(preorder)
inorder_lookup = {n: i for i, n in enumerate(inorder)}
def helper(start, end):
if start > end:
return None
root_val = next(preorder_iterator)
root = TreeNode(root_val)
idx = inorder_lookup[root_val]
root.left = helper(start, idx-1)
root.right = helper(idx+1, end)
return root
return helper(0, len(inorder)-1)
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