题目
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.)
解题思路
代码
# 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, preorder: List[int]) -> TreeNode:
def dfs(l, r):
if l == r:
return TreeNode(preorder[l])
if l > r:
return None
val = preorder[l]
node = TreeNode(val)
pos = l + 1
while pos < self.n and preorder[pos] < val:
pos += 1
node.left = dfs(l + 1, pos - 1)
node.right = dfs(pos, r)
return node
self.n = len(preorder)
return dfs(0, self.n - 1)
