Leetcode 95 不同的二叉搜索树 II

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    // 最后需要返回一个所有根节点的集合， 所以这个题思路有点怪
    // 因为每次深搜返回左、右子树的根节点集合，所以还需要它动将他们拼接
    public List<TreeNode> dfs(int left, int right){
      List<TreeNode> allNodes = new ArrayList<>();
      if(left > right){
        allNodes.add(null);
        return allNodes;
      }
      for(int i = left; i <= right; i++){
        List<TreeNode> leftTrees = dfs(left, i - 1);
        List<TreeNode> rightTrees = dfs(i + 1, right);
        for(TreeNode leftNode : leftTrees){
          for(TreeNode rightNode : rightTrees){
            TreeNode now = new TreeNode(i);
            now.left = leftNode;
            now.right = rightNode;
            allNodes.add(now);
          }
        }
      }
      return allNodes;
    }
    public List<TreeNode> generateTrees(int n) {
      if(n == 0){
        return new ArrayList<TreeNode>();
      }
      return dfs(1, n);
    }
}