原题地址：https://oj.leetcode.com/problems/unique-binary-search-trees-ii/
题意：接上一题，这题要求返回的是所有符合条件的二叉查找树，而上一题要求的是符合条件的二叉查找树的棵数，我们上一题提过，求个数一般思路是动态规划，而枚举的话，我们就考虑dfs了。dfs(start, end)函数返回以start，start+1，...，end为根的二叉查找树。
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: # @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为根节点的值，从start遍历到end LeftTree = self.dfs(start, rootval-1) RightTree = self.dfs(rootval+1, end) for i in LeftTree: #i遍历符合条件的左子树 for j in RightTree: #j遍历符合条件的右子树 root = TreeNode(rootval) root.left = i root.right = j res.append(root) return res def generateTrees(self, n): return self.dfs(1, n)
[leetcode]Unique Binary Search Trees II @ Python,布布扣,bubuko.com
[leetcode]Unique Binary Search Trees II @ Python
原文：http://www.cnblogs.com/zuoyuan/p/3752428.html