Given n, generate all structurally unique BST's (binary search trees) that store values 1...n.java
For example,node
Given n = 3, your program should return all 5 unique BST's shown below.git
1 3 3 2 1github
\ / / / \ \ide
3 2 1 1 3 2spa
/ / \ \code
2 1 2 3blog
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.遞歸
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SOLUTION 1:
使用遞歸來作。
1. 先定義遞歸的參數爲左邊界、右邊界,即1到n.
2. 考慮從left, 到right 這n個數字中選取一個做爲根,餘下的使用遞歸來構造左右子樹。
3. 當無解時,應該返回一個null樹,這樣構造樹的時候,咱們會比較方便,不會出現左邊解爲空,或是右邊解爲空的狀況。
4. 若是說左子樹有n種組合,右子樹有m種組合,那最終的組合數就是n*m. 把這全部的組合組裝起來便可
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; left = null; right = null; } 8 * } 9 */ 10 public class Solution { 11 public List<TreeNode> generateTrees(int n) { 12 // 0.07 13 return dfs(1, n); 14 } 15 16 public List<TreeNode> dfs(int left, int right) { 17 List<TreeNode> ret = new ArrayList<TreeNode>(); 18 19 // The base case; 20 if (left > right) { 21 ret.add(null); 22 return ret; 23 } 24 25 for (int i = left; i <= right; i++) { 26 List<TreeNode> lTree = dfs(left, i - 1); 27 List<TreeNode> rTree = dfs(i + 1, right); 28 for (TreeNode nodeL: lTree) { 29 for (TreeNode nodeR: rTree) { 30 TreeNode root = new TreeNode(i); 31 root.left = nodeL; 32 root.right = nodeR; 33 ret.add(root); 34 } 35 } 36 } 37 38 return ret; 39 } 40 }
CODE: