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95.java
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95.java
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__________________________________________________________________________________________________
sample 1 ms submission
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<TreeNode> generateTrees(int n) {
//在循环中调用递归函数求解子问题。思路是每次一次选取一个结点为根,然后递归求解左右子树的所有结果,最后根据左右子树的返回的所有子树,依次选取然后接上(每个左边的子树跟所有右边的子树匹配,而每个右边的子树也要跟所有的左边子树匹配,总共有左右子树数量的乘积种情况),构造好之后作为当前树的结果返回。
List<TreeNode> list=new ArrayList<TreeNode>();
if(n<1) return list;
return helper(1,n);
}
private ArrayList<TreeNode> helper(int left, int right){
ArrayList<TreeNode> res = new ArrayList<TreeNode>();
if (left > right){
res.add(null);
return res;
}
for (int i = left; i <= right; i++){
ArrayList<TreeNode> lefts = helper(left, i-1);//以i作为根节点,左子树由[1,i-1]构成
ArrayList<TreeNode> rights = helper(i+1, right);//右子树由[i+1, n]构成
for (int j = 0; j < lefts.size(); j++){
for (int k = 0; k < rights.size(); k++){
TreeNode root = new TreeNode(i);
root.left = lefts.get(j);
root.right = rights.get(k);
res.add(root);//存储所有可能行
}
}
}
return res;
}
}
__________________________________________________________________________________________________
sample 37000 kb submission
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<TreeNode> generateTrees(int n) {
if(n==0)
return new ArrayList<TreeNode>();
return generate(1,n);
}
private List<TreeNode> generate(int start, int end)
{
List<TreeNode> ans = new ArrayList();
if(start>end)
{
ans.add(null);
return ans;
}
if(start==end)
{
ans.add(new TreeNode(start));
return ans;
}
for(int i=start; i<=end; i++)
{
List<TreeNode> left = generate(start,i-1);
List<TreeNode> right = generate(i+1, end);
for(TreeNode l : left)
{
for(TreeNode r : right)
{
TreeNode node = new TreeNode(i);
node.left = l;
node .right = r;
ans.add(node);
}
}
}
return ans;
}
}
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