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by Sumit Chourasia | Oct 07, 2020 | Category :coding | Tags : algorithm binary-tree data-structure easy leetcode tree

Find Mode in Binary Search Tree - Tree - Easy - LeetCode

Find Mode in Binary Search Tree - Tree - Easy - LeetCode

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].

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).

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    int? prev = null;
    int count = 1;
    int max = 0;
    public int[] FindMode(TreeNode root) {
        var list = new List<int>();
        InOrder(root,list);
        return list.ToArray();
    }
    
    private void InOrder(TreeNode root, List<int> list){
        if(root==null){
            return;
        }
        
        InOrder(root.left, list);
        
        if(prev != null){
            if(root.val==prev){
            count++;
            }
            else{
                count = 1;
            } 
        }
        
        
        if(count>max){
            max = count;
            list.Clear();
            list.Add(root.val);
        }
        else if(count == max){
            list.Add(root.val);
        }
        
        prev = root.val;
        
        InOrder(root.right, list);
    }
}

Time Complexity: O(n)

Space Complexity: O(n) // for recursion stack.