LeetCode OJ：Kth Smallest Element in a BST（二叉树中第k个最小的元素）

Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.

Note: 
You may assume k is always valid, 1 ≤ k ≤ BST's total elements.

求二叉树中第k个最小的元素，中序遍历就可以了，具体代码和另一个Binary Tree Iterator差不多其实，这题由于把=写成了==调bug调了好久，细心细心啊啊啊。代码如下：

 /**
  * 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;
         map<TreeNode *, bool> m;
         if(root == NULL) return ;
         s.push(root);
         while(!s.empty()){
             TreeNode * t = s.top();
             if(t->left && !m[t->left]){
                 s.push(t->left);
                 m[t->left] = true;
                 continue;
             }
             s.pop();　　//这里pop
             k--;
             if(k == )
                 return t->val;
             if(t->right && !m[t->right]){
                 s.push(t->right);
                 m[t->right] = true;
             }

         }
     }
 };

java 版本的如下所示，同儿茶搜索树迭代器实际上是一样的：

 public class Solution {
     public int kthSmallest(TreeNode root, int k) {
         int res;
         HashMap<TreeNode, Integer> map = new HashMap<TreeNode, Integer>();
         Stack<TreeNode> stack = new Stack<TreeNode>();
         if(root == null)
             return 0;
         stack.push(root);
         while(!stack.isEmpty()){
             TreeNode node = stack.peek();
             while(node.left != null && !map.containsKey(node.left)){
                 stack.push(node.left);
                 map.put(node.left, 1);
                 node = node.left;
             }
             if(--k == 0)
                 return node.val;
             stack.pop(); //这一步不要忘了
             if(node.right != null && !map.containsKey(node.right)){
                 stack.push(node.right);
                 map.put(node.right, 1);
             }
         }
         return 0;
     }
 }