day19

1.剑指 Offer 64. 求 1 + 2 + … + n

1 class Solution {
2 public:
3     int sumNums(int n) {
4        n > 1 && (n += sumNums(n - 1));
5        return n;
6     }
7 };

2.剑指 Offer 68 - I. 二叉搜索树的最近公共祖先

 因为是二叉搜索树，所以根据其性质可分为三种情况（使p < q）：

  i.p、q一个在左子树一个在右子树，或者 其中一个等于root，公共祖先就会是root

  ii.q的值小于root，公共祖先会在左子树

  iii.p的值大于root，公共祖先会在右子树

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
13         if(p -> val > q -> val) return lowestCommonAncestor(root,q,p);
14         if(p -> val == root -> val || q -> val == root -> val || (root -> val > p -> val && root -> val < q -> val)) return root;
15         if(q -> val < root -> val) return lowestCommonAncestor(root -> left,p,q);
16         /*else if(p -> val > root -> val)*/
17         return lowestCommonAncestor(root -> right,p,q);
18     }
19 };

3.剑指 Offer 68 - II. 二叉树的最近公共祖先

 思路与二叉搜索树那里一样，只是二叉搜索树可根据其左右节点值大小直接判断p、q是否在左右子树，而此题要遍历其左右子树才可判断（find函数）

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
13         if(root -> val == p -> val || root -> val == q -> val )  
14         return root;
15         if(find(root -> left,p -> val) && find(root -> left,q -> val))
16          return lowestCommonAncestor(root -> left,p,q);
17         else if(find(root -> right,p -> val) && find(root -> right,q -> val))
18          return lowestCommonAncestor(root -> right,p,q);
19         return root;//剩下一种情况就是p、q分别在左右子树上
20     }
21 
22     bool find(TreeNode* root,int val){
23         if(root == nullptr) return false;
24         if(root -> val == val) return true;
25         if(find(root -> left,val) || find(root -> right,val)) return true;
26         return false; 
27     }
28 };

 理解  lowestCommonAncestor函数返回的是最近公共祖先

 left不为空说明左子树存在最近公共祖先，right同理

 4种情况：i.left != nullptr && right != nullptr，说明p、q一个在左子树一个在右子树，即最近公共祖先为root

                 ii.left == nullptr && right == nullptr,说明 rootroot 的左右子树中都不包含 p,q ，返回空

                 iii.left != nullptr && right == nullptr,说明p、q最近公共祖先在左子树，返回left

                 iv.left == nullptr && right != nullptr,说明p、q最近公共祖先在右子树，返回right

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
13         if(root == nullptr || root == p || root == q) return root;//这里,root为空的话返回的也是nullptr
14         TreeNode* left = lowestCommonAncestor(root -> left,p,q);
15         TreeNode* right = lowestCommonAncestor(root -> right,p,q);
16         if(left != nullptr && right != nullptr)  return root;
17         if(left == nullptr && right == nullptr)  return nullptr;
18         if(left) return left;
19         /* if(right) */
20         return right;
21     }
22 };

 

 
posted @ 2022-07-15 22:13  balabalahhh  阅读(16)  评论(0)    收藏  举报