Unique Binary Search Trees II 输出二叉树的所有组合@LeetCode

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总结一点，要求出所有组合的问题，一般用dfs；如果要求总数，用dp。

另外，这道题的思路就是对1..n中的每一个数都依次让它做root，然后分出左右区间，再递归求解。最后把左右区间求得的子结果依次分别作为root的左右孩子，因此总共要3次循环。

还有值得注意的技巧是，当begin>end时，要往ret AL里面添加null，使得每个AL里面至少有一个元素（null）。这样可以避免判断只有左区间或只有右区间的情况。

package Level4;import java.util.ArrayList;import Utility.TreeNode;/** * Unique Binary Search Trees II  *  * Given n, generate all structurally unique BST's (binary search trees) that store values 1...n.For example,Given n = 3, your program should return all 5 unique BST's shown below.   1         3     3      2      1    \       /     /      / \      \     3     2     1      1   3      2    /     /       \                 \   2     1         2                 3confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ. */public class S95 {public static void main(String[] args) {ArrayList<TreeNode> ret = generateTrees(3);for (TreeNode treeNode : ret) {treeNode.print();System.out.println();}}public static ArrayList<TreeNode> generateTrees(int n) {return dfs(0, n-1);}public static ArrayList<TreeNode> dfs(int begin, int end){ArrayList<TreeNode> ret = new ArrayList<TreeNode>();if(begin > end){ret.add(null);// 相当于占位符return ret;}for(int i=begin; i<=end; i++){ArrayList<TreeNode> left = dfs(begin, i-1);// left和right至少会有一个元素null！ArrayList<TreeNode> right = dfs(i+1, end);for(int j=0; j<left.size(); j++){// 因为size至少为1，所以左右都会被访问到for(int k=0; k<right.size(); k++){TreeNode root = new TreeNode(i+1);// root node, i+1 因为第一个从1而不是从0开始root.left = left.get(j);root.right = right.get(k);ret.add(root);}}}return ret;}}

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; left = null; right = null; } * } */public class Solution {    public List<TreeNode> generateTrees(int n) {        return getTree(1, n);    }        public List<TreeNode> getTree(int left, int right) {        List<TreeNode> list = new ArrayList<TreeNode>();        if(left > right) {            list.add(null);            return list;        }                for(int i=left; i<=right; i++) {            List<TreeNode> leftList = getTree(left, i-1);            List<TreeNode> rightList = getTree(i+1, right);            for(int j=0; j<leftList.size(); j++) {                for(int k=0; k<rightList.size(); k++) {                    TreeNode root = new TreeNode(i);                    list.add(root);                    root.left = leftList.get(j);                    root.right = rightList.get(k);                                    }            }        }                return list;    }}

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