501. Find Mode in Binary Search Tree

Given a binary search tree (BST) withduplicates, find all the mode(s) (the most frequently occurred element) in thegiven BST.

Assume a BST is defined as follows:

 The left subtree of a node contains onlynodes with keys less than or equal to the node's key.

The right subtree of a node contains onlynodes with keys greater than or equal to the node's key.

Both the left and right subtrees must alsobe binary search trees.

For example:

Given BST [1,null,2,2],

   1

    \

    2

    /

   2

return [2].

Note: If a tree has more than one mode, youcan return them in any order.

Follow up: Could you do that without usingany extra space? (Assume that the implicit stack space incurred due torecursion does not count).

    翻译：给定具有重复的二叉搜索树（BST），找到给定BST中的所有模式（最频繁出现的元素）。

    假设BST定义如下：节点的左子树仅包含键小于或等于节点键的节点。节点的右子树仅包含密钥大于或等于节点密钥的节点。左和右子树都必须是二叉搜索树。

例如：

给定BST [1，null，2,2]，

    1

     \“

      2

     /

    2

return [2]。

注意：如果树有多个模式，您可以按任何顺序返回它们。

跟进：你能没有使用任何额外的空间吗？ （假设由于递归导致的隐式堆栈空间不计数）。

    分析：求二叉搜索树的节点中那几个出现的次数最后，找到并返回一个整形数组。由于二叉搜索树的中序遍历时一个递增的数列，所有将它进行中序遍历。Max是节点中出现的次数最大值，num表示当前的数值，numValue表示当前数字出现的次数。代码如下：

/**

 *Definition for a binary tree node.

 *public class TreeNode {

 *    int val;

 *    TreeNode left;

 *    TreeNode right;

 *    TreeNode(int x) { val = x; }

 * }

 */

public class Solution {

   List<Integer> ls=new LinkedList();

   int max=0,num=Integer.MIN_VALUE,numValue=0;

   public void orderVisit(TreeNode root){

        if (root == null) return;

        orderVisit(root.left);

   // Visit(root);

       numValue++;

       if (root.val != num) {

           num = root.val;

           numValue = 1;

       }

       if (numValue > max) {

           max = numValue;

           ls.clear();;

           ls.add(root.val);

       } else if (numValue == max) {

           ls.add(root.val);

       }

        orderVisit(root.right);

    }

   public int[] findMode(TreeNode root) {

       //中序遍历后会产生一组数据

       orderVisit(root);

       int []sum=new int[ls.size()];

       for(int i=0;i<ls.size();i++){

           sum[i]=ls.get(i);

       }

       return sum;

  

     }

}