501. Find Modes in Binary Search Tree

Edited by Charles

Easy · Python · Binary Tree

Problem

Read the original problem on LeetCode.

The official page contains the problem statement, examples, and constraints.

Explanation, examples & other approaches

Core idea

Inorder traversal visits BST values in sorted order. Count consecutive equal values and retain every value matching the highest frequency.

Original study notes follow, including intermediate snippets and earlier attempts.

Iteration:

Based on the core idea of the inorder traversal, we need to deal with such cases (the inorder traversal guarantees the list of traversal node values is non-descending. Thus we only need to compare pre and cur to count the modes.):

1. pre is None, that means cur is still going to find the most left node.
2. cur.val = pre.val, that means in the inorder traversal we found duplicates.
3. cur.val ≠ pre.val, that means there won’t be any identical values as pre.val.

For the count, we use maxFreq to record the highest frequency so far.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def findMode(self, root: Optional[TreeNode]) -> List[int]:
        modes = []
        stack = []
        cur = root
        pre = None
        count = 0
        maxFreq = 0

        while cur or stack:
            if cur:
                stack.append(cur)
                cur = cur.left
            else:
                cur = stack.pop()
                if not pre:
                    count = 1
                elif cur.val == pre.val:
                    count += 1
                else:
                    count = 1

                if count == maxFreq:
                    modes.append(cur.val)
                if count > maxFreq:
                    modes = [cur.val]
                    maxFreq = count

                pre = cur
                cur = cur.right

        return modes

Study record

Field	Value
Attempts	1
Mastery	❚
The Latest Attempt	January 16, 2026

Python solution

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def __init__(self):
        self.modes = []
        self.maxFreq = 0
        self.count = 0
        self.prev = None

    def findMode(self, root: Optional[TreeNode]) -> List[int]:
        self.traverse(root)
        return self.modes

    def traverse(self, node):
        if not node:
            return

        self.traverse(node.left)

        if not self.prev:
            self.count = 1
        elif node.val == self.prev.val:
            self.count += 1
        else:
            self.count = 1

        if self.count > self.maxFreq:
            self.modes = [node.val]
            self.maxFreq = self.count
        elif self.count == self.maxFreq:
            self.modes.append(node.val)

        self.prev = node

        self.traverse(node.right)

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