230. 二叉搜索树中第K小的元素 

 1 class Solution 
 2 {
 3 public:
 4     int kthSmallest(TreeNode* root, int k) 
 5     {
 6         stack<TreeNode*> s;
 7         TreeNode *cur = root;
 8         while(!s.empty() || cur)
 9         {
10             if(cur)
11             {
12                 s.push(cur);
13                 cur = cur->left;
14             }
15             else
16             {
17                 cur = s.top();
18                 s.pop();
19                 k--;
20                 if(0 == k) return cur->val;
21                 cur = cur->right;
22             }
23         }
24         return 0;
25     }
26 };

 

 1 // left = 左子树节点的个数
 2 
 3 //1、if k <= left: return dfs(root->left,k);
 4 //2、if k == left + 1: return root->val;
 5 //3、if k > left + 1: return dfs(root->right,k - left - 1);
 6 
 7 class Solution 
 8 {
 9 public:
10     int kthSmallest(TreeNode* root, int k) 
11     {
12         int left = findChild(root->left);
13 
14         if(k <= left) return kthSmallest(root->left,k);
15         else if(k == left + 1) return root->val;
16         return kthSmallest(root->right,k - left - 1);
17     }
18 
19     int findChild(TreeNode* root)//记录当前节点及其左右节点的个数
20     {
21         if(!root) return 0;
22         return findChild(root->left) + findChild(root->right) + 1;
23     }
24 };

 

posted @ 2020-04-12 13:58  Jinxiaobo0509  阅读(183)  评论(0)    收藏  举报