Unique Binary Search Trees II *

Given an integern, generate all structurally uniqueBST's(binary search trees) that store values 1...n.

For example, Givenn= 3, your program should return all 5 unique BST's shown below.

   1         3     3      2      1
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3

Solution:

    if start > end:

        return \[None\]  非常重要，因为如果start > end 说明没有任何所以需要补入Noned的节点进入list 这样才可以补空位。

    常犯的错误，看见start end, 就直接二分法了，应该还是从start 开始一个一个的遍历。

double for means Left * Right

DFS + 分治

class Solution(object):
    def generateTrees(self, n):
        """
        :type n: int
        :rtype: List[TreeNode]
        """
        if n <= 0 :
            return []
        return self.dfs(1, n)

    def dfs(self, start, end):
        if start > end: return [None]
        res = []
        for rootval in range(start, end+1):
            LeftTree = self.dfs(start, rootval-1)
            RightTree = self.dfs(rootval+1, end)
            for i in LeftTree:
                for j in RightTree:
                    root = TreeNode(rootval)
                    root.left = i
                    root.right = j
                    res.append(root)
        return res