二叉搜索树的第k个节点

一是利用大家都能想到的中序遍历递归法：

public class Solution {
    List<TreeNode> list = new ArrayList<>();
    TreeNode KthNode(TreeNode pRoot, int k)
    {
        if(pRoot == null || k < 1){
            return null;
        }
        minOrder(pRoot, k);
        //当k超过节点个数时
        if(list.size() < k)
            return null;
        return list.get(k-1);
    }

    void minOrder(TreeNode pRoot, int k){
        if(pRoot != null){
            minOrder(pRoot.left, k);
            list.add(pRoot);
            k = k - 1;
            if(k == 0)
                return;
            minOrder(pRoot.right, k);
        }
    }
}

public class Solution {
    List<TreeNode> list = new ArrayList<>();
    TreeNode KthNode(TreeNode pRoot, int k)
    {
        if(pRoot == null || k < 1){
            return null;
        }
        TreeNode node = pRoot;
        int count = 0;
        Stack<TreeNode> stack = new Stack<>();

        while(!stack.isEmpty() || node != null){
            //中序遍历
            //遍历到左子树为空
            if(node != null){
                stack.push(node);
                node = node.left;
            }else{
                //再弹出去
                node = stack.pop();
                count++;
                //当弹到第k个时，直接返回
                if(count == k)
                    return node;
                //否则，遍历右子树
                node = node.right;
            }
        }
        return null;
    }

}