• 1. 给表达式加括号
• 2. 不同的二叉搜索树

1. 给表达式加括号

241. Different Ways to Add Parentheses (Medium)

Leetcode / 力扣

// html
Input: "2-1-1".

((2-1)-1) = 0
(2-(1-1)) = 2

Output : [0, 2]

// java
public List diffWaysToCompute(String input) {
    List ways = new ArrayList<>();
    for (int i = 0; i < input.length(); i++) {
        char c = input.charAt(i);
        if (c == '+' || c == '-' || c == '*') {
            List left = diffWaysToCompute(input.substring(0, i));
            List right = diffWaysToCompute(input.substring(i + 1));
            for (int l : left) {
                for (int r : right) {
                    switch (c) {
                        case '+':
                            ways.add(l + r);
                            break;
                        case '-':
                            ways.add(l - r);
                            break;
                        case '*':
                            ways.add(l * r);
                            break;
                    }
                }
            }
        }
    }
    if (ways.size() == 0) {
        ways.add(Integer.valueOf(input));
    }
    return ways;
}

2. 不同的二叉搜索树

95. Unique Binary Search Trees II (Medium)

Leetcode / 力扣

给定一个数字 n，要求生成所有值为 1...n 的二叉搜索树。

// html
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

// java
public List generateTrees(int n) {
    if (n < 1) {
        return new LinkedList();
    }
    return generateSubtrees(1, n);
}

private List generateSubtrees(int s, int e) {
    List res = new LinkedList();
    if (s > e) {
        res.add(null);
        return res;
    }
    for (int i = s; i <= e;="" ++i)="" {="" list leftSubtrees = generateSubtrees(s, i - 1);
        List rightSubtrees = generateSubtrees(i + 1, e);
        for (TreeNode left : leftSubtrees) {
            for (TreeNode right : rightSubtrees) {
                TreeNode root = new TreeNode(i);
                root.left = left;
                root.right = right;
                res.add(root);
            }
        }
    }
    return res;
}