题解 | #二叉搜索树的第k个节点#

中序遍历

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param proot TreeNode类 
     * @param k int整型 
     * @return int整型
     */
    int KthNode(TreeNode* proot, int k) {
      if (proot == nullptr) {
        return -1;
      }
      
      int count = 0;
      int res = -1;
      
      curision(proot, k, count, res);
      
      return res;
    }
  private:
   void curision(TreeNode *root, const int &k, int &count, int &res) {
     if (root == nullptr) {
       return ;
     }
    
     curision(root->left, k, count, res);
     
     ++count;
     
     if (count == k) {
       res = root->val;
       return ;
     }
     
     curision(root->right, k, count, res);
   }
};