501. Find Mode in Binary Search Tree

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

Assume a BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than or equal to the node's key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
• Both the left and right subtrees must also be binary search trees.

 

For example:
Given BST [1,null,2,2],

return [2].

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

Follow up: Could you do that without using any extra space? (Assume that the implicit stack space incurred due to recursion does not count).

 

Solution1：//找出出现频率最高的元素

class Solution {

    Integer prev = null;

    int count = 1;

    int max = 0;

    public int[] findMode(TreeNode root) {

        if (root == null) return new int[0];

        List<Integer> list = new ArrayList<>();  //创建一个ArrayList，因为是动态大小的，所以方便数据添加。

        traverse(root, list);

        int[] res = new int[list.size()];

        for (int i = 0; i < list.size(); i++) res[i] = list.get(i);

        return res; 

    }

 

    private void traverse(TreeNode root, List list) {

         if (root == null) return;

         traverse(root.left, list);  

         if (prev != null) {  //在第一次使用traverse方法时，prev是null所以此判断语句不会运行，之后运行到此处                                           时，prev不是null所以会根据情况判断是否把count加一

             if (root.val == prev)

                 count ++;

             else 

                 count = 1;

         }

         if (count > max) { //如果此节点值的数量大于max，则把此节点加入ArrayList中（并删了之前的节点）

             max = count;

             list.clear();

             list.add(root.val);

         } else if (count == max) {

             list.add(root.val);

         }

         prev = root.val;  //最后才把节点的值赋给prev

         traverse(root.right, list);

    }

}