leetCode 95.Unique Binary Search Trees II （唯一二叉搜索树） 解题思路和方法

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                 3

confused what "{1,#,2,3}" means? >
read more on how binary tree is serialized on OJ.

思路：这一题尽管比較繁琐。可是难度不算非常大，能够运用排列组合的思想，全排列，然后查找符合要求的二叉搜索树就可以。

也能够运用递归，将数分为左右子树，进而简化求解。

排列组合思想代码例如以下（不知为什么OJ未通过，n=2时报错。但本地測试全然正确）：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    boolean[] b;
    List<TreeNode> list;
    Set<String> set = new HashSet<String>();
    public List<TreeNode> generateTrees(int n) {
        b = new boolean[n];
        int[] a = new int[n];
        list = new ArrayList<TreeNode>();
        for(int i = 0; i < n; i++){
            a[i] = i+1;
        }
        create(a,null,n);
        return list;
    }
 
    /**
     * 生成二叉搜索树
     */
    private void create(int[] a,TreeNode root,int nn){
        if(nn == 0){
        	TreeNode q = root;
        	String s = preOrder(q, "");
        	//System.out.println(s);
        	if(set.add(s)){
        		list.add(root);
        	}
            return;
        }
 
        for(int i = 0; i < a.length; i++){
            if(!b[i]){
                b[i] = true;
                TreeNode p = new TreeNode(a[i]);
                root = insert(root,p);
                create(a,root,nn-1);
                root = delete(root,p);
                b[i] = false;
            }
        }
    }
 
    /**
     * 前序遍历
     * @param root
     * @param s
     * @return
     */
   private String preOrder(TreeNode root,String s){
	   if(root != null){
		   s += root.val;
 
		   if(root.left != null){
			   s = preOrder(root.left, s);
		   }
 
		   if(root.right != null){
			   s = preOrder(root.right, s);
		   }
	   }
	return s;
   }
 
    /**
     * 删除节点
     * @param root
     * @param p
     * @return
     */
    private TreeNode delete(TreeNode root, TreeNode p) {
    	if(root.val == p.val)
    		return null;
    	if(root.val < p.val){
    		root.right = delete(root.right,p);
    	}else{
    		root.left = delete(root.left,p);
    	}
		return root;
	}
 
	/**
     * 将新节点插入二叉搜索树
     */
    private TreeNode insert(TreeNode root,TreeNode node){
    	TreeNode p = root;
 
        if(p == null){
            p = node;
            return p;
        }
        if(node.val < p.val){
            root.left = insert(p.left,node);
        }else{
            root.right = insert(p.right,node);
        }
        return root;
    }
}

递归解法：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
	public List<TreeNode> generateTrees(int n){
		List<TreeNode> list = new ArrayList<TreeNode>();
		if(n <= 0){
			list.add(null);
			return list;
		}
		list = createTree(1,n);
 
		return list;
	}
	/**
	 * 循环生产二叉搜索树
	 * @param i 開始值
	 * @param j 结束值
	 * @return
	 */
	private List<TreeNode> createTree(int i, int j){
 
		List<TreeNode> list = new ArrayList<TreeNode>();
		//起始大于结束值，加入null
		if(i > j){
			list.add(null);
			return list;
		}
		//相等也即加入一个
		if(i == j){
			list.add(new TreeNode(i));
			return list;
		}
		//循环加入
		for(int k = i; k <= j; k++){
			//左子树肯定比i小
			List<TreeNode> left = createTree(i,k-1);
			//右子树肯定比i大
			List<TreeNode> right = createTree(k+1,j);
			//将结果循环加入
			for(TreeNode l:left){
				for(TreeNode r:right){
					TreeNode root = new TreeNode(k);
					root.left = l;
					root.right = r;
					list.add(root);
				}
			}
		}
		return list;
	}
}