Given an array of integers preorder, which represents the preorder traversal of a BST (i.e., binary search tree), construct the tree and return its root.
It is guaranteed that there is always possible to find a binary search tree with the given requirements for the given test cases.
A binary search tree is a binary tree where for every node, any descendant of Node.left has a value strictly less than Node.val, and any descendant of Node.right has a value strictly greater than Node.val.
A preorder traversal of a binary tree displays the value of the node first, then traverses Node.left, then traverses Node.right.
Example 1:
Input: preorder = [8,5,1,7,10,12]
Output: [8,5,10,1,7,null,12]
Example 2:
Input: preorder = [1,3]
Output: [1,null,3]
Constraints:
-
1 <= preorder.length <= 100 -
1 <= preorder[i] <= 1000 - All the values of
preorderare unique.
SOLUTION:
# 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 getRoot(self, p, q):
return min((i for i in range(p, q)), key = lambda x: self.valIndex[self.inorder[x]])
def buildTreeRec(self, p, q):
if p < q:
root = TreeNode()
currRoot = self.getRoot(p, q)
root.val = self.inorder[currRoot]
if currRoot > p:
root.left = TreeNode()
root.left = self.buildTreeRec(p, currRoot)
if currRoot < q - 1:
root.right = TreeNode()
root.right = self.buildTreeRec(currRoot + 1, q)
return root
def bstFromPreorder(self, preorder: List[int]) -> Optional[TreeNode]:
n = len(preorder)
self.inorder = sorted(preorder)
self.valIndex = {}
for i, item in enumerate(preorder):
self.valIndex[item] = i
return self.buildTreeRec(0, n)
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