Count of nodes that are greater than Ancestors

Given the root of a tree, the task is to find the count of nodes which are greater than all of its ancestors.
Examples:

Input: 
  4
 / \
5   2
   / \
  3   6
Output: 3
The nodes are 4, 5 and 6.

Input: 
   10
  /  \
 8    6
  \    \
   3    5
  /
 1
Output: 1

Approach: The problem can be solved using dfs. In every function call, pass a variable maxx which stores the maximum among all the nodes traversed so far, and every node whose value is greater than maxx is the node that satisfies the given condition. Hence, increment the count.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Structure for the node of the tree
struct Tree {
    int val;
    Tree* left;
    Tree* right;
    Tree(int _val)
    {
        val = _val;
        left = NULL;
        right = NULL;
    }
};

// Dfs Function
void dfs(Tree* node, int maxx, int& count)
{
    // Base case
    if (node == NULL) {
        return;
    }
    else {

        // Increment the count if the current
        // node's value is greater than the
        // maximum value in it's ancestors
        if (node->val > maxx)
            count++;

        // Left traversal
        dfs(node->left, max(maxx, node->val), count);

        // Right traversal
        dfs(node->right, max(maxx, node->val), count);
    }
}

// Driver code
int main()
{

    Tree* root = new Tree(4);
    root->left = new Tree(5);
    root->right = new Tree(2);
    root->right->left = new Tree(3);
    root->right->right = new Tree(6);

    // To store the required count
    int count = 0;

    dfs(root, INT_MIN, count);

    cout << count;

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

static int count;

// Structure for the node of the tree
static class Tree
{
    int val;
    Tree left;
    Tree right;
    Tree(int _val)
    {
        val = _val;
        left = null;
        right = null;
    }
};

// Dfs Function
static void dfs(Tree node, int maxx)
{
    // Base case
    if (node == null)
    {
        return;
    }
    else
    {

        // Increment the count if the current
        // node's value is greater than the
        // maximum value in it's ancestors
        if (node.val > maxx)
            count++;

        // Left traversal
        dfs(node.left, Math.max(maxx, node.val));

        // Right traversal
        dfs(node.right, Math.max(maxx, node.val));
    }
}

// Driver code
public static void main(String[] args)
{
    Tree root = new Tree(4);
    root.left = new Tree(5);
    root.right = new Tree(2);
    root.right.left = new Tree(3);
    root.right.right = new Tree(6);

    // To store the required count
    count = 0;

    dfs(root, Integer.MIN_VALUE);

    System.out.print(count);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program for the
# above approach
from collections import deque

# A Tree node
class Tree:

    def __init__(self, x):

        self.val = x
        self.left = None
        self.right = None

count = 0

# Dfs Function
def dfs(node, maxx):

    global count

    # Base case
    if (node == None):
        return
    else:

        # Increment the count if
        # the current node's value
        # is greater than the maximum
        # value in it's ancestors
        if (node.val > maxx):
            count += 1

        # Left traversal
        dfs(node.left,
            max(maxx,
                node.val))

        # Right traversal
        dfs(node.right,
            max(maxx,
                node.val))

# Driver code
if __name__ == '__main__':

    root = Tree(4)
    root.left = Tree(5)
    root.right = Tree(2)
    root.right.left = Tree(3)
    root.right.right = Tree(6)

    # To store the required
    # count
    count = 0

    dfs(root,
        -10 ** 9)
    print(count)

# This code is contributed by Mohit Kumar 29

C#

// C# implementation of the approach
using System;

class GFG
{
static int count;

// Structure for the node of the tree
public class Tree
{
    public int val;
    public Tree left;
    public Tree right;
    public Tree(int _val)
    {
        val = _val;
        left = null;
        right = null;
    }
};

// Dfs Function
static void dfs(Tree node, int maxx)
{
    // Base case
    if (node == null)
    {
        return;
    }
    else
    {

        // Increment the count if the current
        // node's value is greater than the
        // maximum value in it's ancestors
        if (node.val > maxx)
            count++;

        // Left traversal
        dfs(node.left, Math.Max(maxx, node.val));

        // Right traversal
        dfs(node.right, Math.Max(maxx, node.val));
    }
}

// Driver code
public static void Main(String[] args)
{
    Tree root = new Tree(4);
    root.left = new Tree(5);
    root.right = new Tree(2);
    root.right.left = new Tree(3);
    root.right.right = new Tree(6);

    // To store the required count
    count = 0;

    dfs(root, int.MinValue);

    Console.Write(count);
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>
// Javascript implementation of the approach

let count=0;
// Structure for the node of the tree
class Tree
{
    constructor(val)
    {
        this.val=val;
        this.left=this.right=null;
    }
}

// Dfs Function
function dfs(node,maxx)
{
    // Base case
    if (node == null)
    {
        return;
    }
    else
    {

        // Increment the count if the current
        // node's value is greater than the
        // maximum value in it's ancestors
        if (node.val > maxx)
            count++;

        // Left traversal
        dfs(node.left, Math.max(maxx, node.val));

        // Right traversal
        dfs(node.right, Math.max(maxx, node.val));
    }
}

// Driver code
let root = new Tree(4);
root.left = new Tree(5);
root.right = new Tree(2);
root.right.left = new Tree(3);
root.right.right = new Tree(6);

// To store the required count
count = 0;

dfs(root, Number.MIN_VALUE);

document.write(count);


// This code is contributed by unknown2108
</script>

Output

3

Time Complexity: O(N) where N is the number of nodes in given binary tree.

Space Complexity: O(h) where h is the height of binary tree due to stack recursion call.