原題地址:https://oj.leetcode.com/problems/unique-binary-search-trees-ii/
題意:接上一題,這題要求返回的是所有符合條件的二叉查找樹,而上一題要求的是符合條件的二叉查找樹的棵數,我們上一題提過,求個數一般思路是動態規划,而枚舉的話,我們就考慮dfs了。dfs(start, end)函數返回以start,start+1,...,end為根的二叉查找樹。
代碼:
# Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: # @return a list of tree node def dfs(self, start, end): if start > end: return [None] res = [] for rootval in range(start, end+1): #rootval為根節點的值,從start遍歷到end LeftTree = self.dfs(start, rootval-1) RightTree = self.dfs(rootval+1, end) for i in LeftTree: #i遍歷符合條件的左子樹 for j in RightTree: #j遍歷符合條件的右子樹 root = TreeNode(rootval) root.left = i root.right = j res.append(root) return res def generateTrees(self, n): return self.dfs(1, n)