Construct Binary Search Tree from Preorder Traversal

Return the root node of a binary search tree that matches the given preorder traversal.

(Recall that a binary search tree is a binary tree where for every node, any descendant of node.left has a value < node.val, and any descendant of node.right has a value > node.val. Also recall that a preorder traversal displays the value of the node first, then traverses node.left, then traverses node.right.)

Example 1:

Input: [8,5,1,7,10,12]
Output: [8,5,10,1,7,null,12]

分析

因为bst 所以左树《root.val<右树

按照范围，左边第一个比max小的数字就是root，所以设置max，顺序找第一个数字 《=max，然后递归

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def bstFromPreorder(self, p: List[int]) -> TreeNode:
        
        def helper(stop):
            if p and p[0] <= stop:
                root = TreeNode(p.pop(0))
                root.left = helper(root.val)
                root.right = helper(stop)
                return root
            
        #p.reverse()   
        return helper(max(p))
            
            
        

recursive

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def bstFromPreorder(self, p: List[int]) -> TreeNode:
        if not p:
            return None
        if len(p) == 1:
            return TreeNode(p[0])
        root = TreeNode(p[0])
        i=1
        while i<len(p) and p[i]<root.val:
            i+=1
        root.left = self.bstFromPreorder(p[1:i])
        root.right = self.bstFromPreorder(p[i:])
        return root