【leetcode】501. Find Mode in Binary Search Tree-优快云博客

本文介绍了一种在含有重复元素的二叉搜索树中查找最频繁出现元素的方法，利用递归进行中序遍历，通过比较节点值来更新计数并找到众数。

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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).

思考：又是一道与二叉搜索树相关的题，对于该题我们可以反应到可以使用递归来实现。由于该道题的定义二叉搜索树的左子树结点的值小于或者等于根结点的值，右子树结点大于或者等于根结点的值，所以可能会出现根结点，左右子树结点的值都相等的情况。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    Integer prev = null;
    int count = 1;
    int max = 0;
    
    public int[] findMode(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        if(root==null){
            return new int[0];
        }
        traverseInorder(list,root);
        int[] res = new int[list.size()];
        for(int i=0;i<list.size();i++){
            res[i] = list.get(i);
        }
        return res;
    }
    
    public void traverseInorder(List<Integer> list,TreeNode root){
        if(root==null){
            return;
        }
        //先遍历左子树，获取prev值
        traverseInorder(list,root.left);
        //判断prev变量是否有值
        if(prev != null){
            if(prev == root.val){
                count++;
            } else {
                count = 1;
            }
        }
        if(count > max){
            max = count;
            list.clear();
            list.add(root.val);
        } else if(count==max){
            list.add(root.val);
        } 
        //记录prev值
        prev = root.val;
        //再遍历右子树
        traverseInorder(list,root.right);
    }
}