LeetCode: Unique Binary Search Trees [095]

【题目】

Given n, how many structurally unique BST's (binary search trees) that store values 1...n?

For example,
Given n = 3, there are a total of 5 unique BST's.

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

【题意】

给定一个数字n, 问用1,2,3,4,5...n这n个值，能构造多少棵合法的二叉搜索树

【思路】

对于给定[1,n]区间，先确定全部可能的根。如果根为k, 则该二叉树左子树的取值区间为[1, k-1]和右子树的取值区间[k+1, n]。
而二叉搜索搜索树的随意一个节点的左右子树也是二叉搜索树，因此我们须要确定[1,k-1]和[k+1, n]上构造的二叉搜索树的数目, 比方分别为left[k], right[k]。

则以k的根的二叉搜索树的数目即为，left[k]*right[k]

本题用递归来解决。

【代码】

class Solution {
public:int binaryTreeNums(int start, int end){//[start, end]区间上构造二叉树的数目if(start>=end)return 1;     //start<end表示空子树， start==end表示叶子节点int treeNums=0;for(int root=start; root<=end; root++){int leftCount = binaryTreeNums(start, root-1);  //计算左子树的数目int rightCount = binaryTreeNums(root+1, end);   //计算右子树的数目treeNums+= leftCount*rightCount;}return treeNums;}int numTrees(int n) {if(n==0)return 0;return binaryTreeNums(1, n);}
};

转载于:https://www.cnblogs.com/mengfanrong/p/5152210.html

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