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],

   1
    \
     2
    /
   2

return [2].

題意

從一個二叉搜尋樹中找到出現最頻繁的元素，即出現次數最多的元素。

思路

採用三個遍歷方式之一；因為要記錄相同元素出現的次數，所以需要記錄上一個節點用來判斷，如果上一個節點與當前節點相同，

則出現次數加1，否則，出現次數重置，當前節點賦值給上一個節點；同時記錄大於等於出現最大次數的元素，所以需要一個數組，

然後需要一個計數器，當如果上一個節點與當前節點不相同時並且當前節點出現次數大於或等於最大出現次數，把該節點值新增到陣列，

並且該計數器加1。大致思路如此

程式碼

public class Solution {
    public int[] findMode(TreeNode root) {
        inorder(root);
        modes = new int[modeCount];
        modeCount = 0;
        currCount = 0;
        inorder(root);
        return modes;
    }
    //記錄元素值
    private int currVal;
    //記錄相同元素出現次數
    private int currCount = 0;
    //記錄出現的最大次數
    private int maxCount = 0;
    //作為modes陣列的下標，記錄出現相同最大次數的元素個數
    private int modeCount = 0;
    
    private int[] modes;
    private void handleValue(int val) {
        if (val != currVal) {
            currVal = val;
            currCount = 0;
        }
        currCount++;
        if (currCount > maxCount) {
            maxCount = currCount;
            modeCount = 1;
        } else if (currCount == maxCount) {
            if (modes != null)
                modes[modeCount] = currVal;
            modeCount++;
        }
    }
    
    private void inorder(TreeNode root) {
        if (root == null) return;
        inorder(root.left);
        handleValue(root.val);
        inorder(root.right);
    }
}