思路:
二叉树搜索树的中序遍历,按照遍历的顺序将元素的值压入容器内,容器中的元素按照从小到大的顺序排列。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int kthSmallest(TreeNode* root, int k) {
stack<TreeNode*> s;
vector<int> v;
TreeNode* p = root;
while (p || !s.empty())
{
while (p)
{
s.push(p);
p = p->left;
}
if (!s.empty())
{
p = s.top();
v.push_back(p->val);
s.pop();
p = p->right;
}
}
return v[k - 1];
}
};
一种更优的解法,不用容器,一旦终须访问到第k个元素直接返回它。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int kthSmallest(TreeNode* root, int k) {
stack<TreeNode*> s;
int n = 0;
TreeNode* p = root;
while (p || !s.empty())
{
while (p)
{
s.push(p);
p = p->left;
}
if (!s.empty())
{
p = s.top();
++n;
if (n == k)
return p->val;
s.pop();
p = p->right;
}
}
return -1;
}
};