96. Unique Binary Search Trees (Tree; DP)

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

 

 法I：后序遍历。同Unique Binary Search Trees II，只是现在要返回的是数量，而无需将子树的所有情况放在数组中返回。

class Solution {
public:
    int numTrees(int n) {
        if(n==) return ;
        else return postOrderTraverse(,n);
    }

    int postOrderTraverse(int start, int end)
    {
        if(start == end){ //递归结束条件：碰到叶子节点（没有子节点）
            return ;
        }

        int leftNum, rightNum, num=;
        for(int i = start; i<=end; i++){ //iterate all roots
            leftNum = ;
            rightNum = ;
            if(i > start){ //count left tree
                leftNum = postOrderTraverse(start, i-);
            }
            if(i < end){ //count right tree
                rightNum = postOrderTraverse(i+, end);
            }

            //visit root: count number for each (leftTree, rightTree) pair
            if(leftNum==) num+=rightNum;
            else if(rightNum==) num+=leftNum;
            else num+=leftNum*rightNum;
        }
        return num;
    }
};

Result:  Time Limit Exceeded

法II：法I超时的原因是进行了很多重复计算。比如，在1-3的子树数量和4-6子树数量是相通的，因为它们的数字跨度相同。

解决方法是用动态规划存储每种数字跨度下的子数数量

class Solution {
public:
    int numTrees(int n) {
        if(n==) return ;
        int dp[n+];
        dp[] = ;
        dp[] = ;
        for(int i = ; i <= n; i++){
            dp[i] = ; //initialize
            for(int j = ; j <= i; j++){ //traverse root
                if(i-j> && j>) //左、右子树都存在
                    dp[i] += dp[j-]*dp[i-j];
                else if(j>) //only left tree
                    dp[i] += dp[j-];
                else //only right tree
                    dp[i] += dp[i-j];
            }
        }
        return dp[n];
    }

};