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 0;
        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 == 0)
                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;
    }
}