Algorithm
Problem Name: 501. Find Mode in Binary Search Tree
Given the root of a binary search tree (BST) with duplicates, return all the mode(s) (i.e., the most frequently occurred element) in it.
If the tree has more than one mode, return them in any order.
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.
Example 1:
Input: root = [1,null,2,2] Output: [2]
Example 2:
Input: root = [0] Output: [0]
Constraints:
- The number of nodes in the tree is in the range
[1, 104]. -105 <= Node.val <= 105
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
const findMode = function(root) {
if(root == null) return []
const hash = {}
traverse(root, hash)
const res = Object.entries(hash).sort((a, b) => b[1] - a[1])
const result = [res[0][0]]
for(let i = 1; i < res.length; i++) {
if(res[i][1] === res[0][1]) result.push(res[i][0])
else break
}
return result
};
function traverse(node, hash) {
if(node === null) return
hash[node.val] = Object.prototype.hasOwnProperty.call(hash, node.val) ? hash[node.val] + 1 : 1
traverse(node.left, hash)
traverse(node.right, hash>
}
Input
root = [1,null,2,2]
Output
[2]
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def findMode(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
from collections import Counter
def traverse(node):
if node: dic[node.val]+=1; traverse(node.left); traverse(node.right)
dic = collections.Counter()
traverse(root)
mx=max(dic.values(),default=0)
return [k for k,v in dic.items() if v==mx]
Input
root = [1,null,2,2]
Output
[2]
#3 Code Example with C# Programming
Code -
C# Programming
using System.Collections.Generic;
namespace LeetCode
{
public class _0501_FindModeInBinarySearchTree
{
private List < int> result = new List<int>();
private int count, modelCount, prevValue;
public int[] FindMode(TreeNode root)
{
InOrder(root);
return result.ToArray();
}
private void InOrder(TreeNode node)
{
if (node == null) return;
InOrder(node.left);
if (prevValue != node.val) count = 1;
else count++;
if (count > modelCount)
{
result.Clear();
result.Add(node.val);
modelCount = count;
}
else if (count == modelCount)
result.Add(node.val);
prevValue = node.val;
InOrder(node.right);
}
}
}
Input
root = [0]
Output
[0]