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0105_Construct_Binary_Tree_from_Preorder_and_Inorder_Traversal.py
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0105_Construct_Binary_Tree_from_Preorder_and_Inorder_Traversal.py
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from typing import List, Optional
class TreeNode: # Definition for a binary tree node.
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def preorder(self):
result = []
def preorderRecursive(node):
if node:
result.append(node.val)
preorderRecursive(node.left)
preorderRecursive(node.right)
preorderRecursive(self)
return result
def inorder(self):
result = []
def inorderRecursive(node):
if node:
inorderRecursive(node.left)
result.append(node.val)
inorderRecursive(node.right)
inorderRecursive(self)
return result
class Solution:
def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
"""
Time: O(N)
Space: O(N)
"""
map_val_pos_inorder = {val: idx for idx, val in enumerate(inorder)}
preorder_index = 0
def process(left, right):
"""
preorder = root, left_preorder, right_preorder
inorder = left_inorder, root, right_inorder
So, if I know where the next root is, I can build the tree from the inorder recursively.
"""
if left > right:
return None
nonlocal preorder_index
value = preorder[preorder_index]
preorder_index += 1
node = TreeNode(value)
node.left = process(left, map_val_pos_inorder[value] - 1)
node.right = process(map_val_pos_inorder[value] + 1, right)
return node
return process(0, len(preorder) - 1)
s = Solution()
preorder_list = [3, 9, 20, 15, 7]
inorder_list = [9, 3, 15, 20, 7]
r = s.buildTree(preorder_list, inorder_list)
print(r.preorder(), preorder_list)
print(r.inorder(), inorder_list)
preorder_list = [1, 2, 3, 4]
inorder_list = [4, 3, 2, 1]
r = s.buildTree(preorder_list, inorder_list)
print(r.preorder(), preorder_list)
print(r.inorder(), inorder_list)