95. Unique Binary Search Trees II(dfs经典题目)

最新推荐文章于 2021-11-05 07:46:42 发布

转载最新推荐文章于 2021-11-05 07:46:42 发布 · 60 阅读

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原文链接：http://www.cnblogs.com/bernieloveslife/p/9744934.html

本文介绍了一种算法，用于生成所有可能的唯一二叉搜索树（BST），这些树存储从1到n的整数值。通过递归地构造左右子树并以1到n的每个值作为根节点，可以得到所有独特的BST组合。

Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1 ... n.

Example:

Input: 3
Output:
[
[1,null,3,2],
[3,2,null,1],
[3,1,null,null,2],
[2,1,3],
[1,null,2,null,3]]
Explanation:
The above output corresponds to the 5 unique BST's shown below:

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

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

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

    def generateTreesDFS(self, left, right):
        if left > right:
            return [None]
        res = []
        for i in range(left, right + 1):
            left_nodes = self.generateTreesDFS(left, i - 1)
            right_nodes = self.generateTreesDFS(i + 1, right)
            for left_node in left_nodes:
                for right_node in right_nodes:
                    root = TreeNode(i)
                    root.left = left_node
                    root.right = right_node
                    res.append(root)
        return res

所有的解就是分别以1,2,3...n做根节点的解的集合.而以i为根节点的解的个数就是左右子树的解的积.
递归求解.

转载于:https://www.cnblogs.com/bernieloveslife/p/9744934.html