[leetcode][tree] Unique Binary Search Trees

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

递归实现（存在大量的重复计算，所以会超时）

class Solution {public:    int numTrees(int n) {if (n <= 0) return 0;if (n <= 2) return n;int res = 0;for (int i = 1; i <= n; ++i){int numTreesOfLeftChild = numTrees(i - 1);int numTreesOfRightChild = numTrees(n - i);if (0 == numTreesOfLeftChild || 0 == numTreesOfRightChild) res += numTreesOfLeftChild + numTreesOfRightChild;else res += numTreesOfLeftChild * numTreesOfRightChild;}return res;}};

动态规划（用table记录）

<pre name="code" class="cpp">class Solution {public:    int numTrees(int n) {int *table = new int[n+1];//table[i]表示i个数能形成的不同二叉树的个数for (int i = 0; i < 3; ++i) table[i] = i;//table初始化for (int i = 3; i <= n; ++i){//求i个数能形成多少种二叉树table[i] = 0;for (int j = 1; j <= i; ++j){//table[i]等于分别由1...i做根节点时可以形成的二叉树的个数的总和int numTreesOfLeftChild = table[j - 1];//左子树节点个数为j-1int numTreesOfRightChild = table[i - j];//右子树节点个数为i-jif (0 == numTreesOfLeftChild || 0 == numTreesOfRightChild) table[i] += numTreesOfLeftChild + numTreesOfRightChild;//当左右子树中有一棵为空时else table[i] += numTreesOfLeftChild * numTreesOfRightChild;//左右子树都不为空时}}int res = table[n];delete[]table;return res;}};