Find Sum of Right Leaves of a Binary Tree in C++

Suppose we have a binary tree we have to find the sum of all right leaves in a given binary tree.

So, if the input is like

then the output will be 17, as there are two right leaves in the binary tree, with values 7 and 10 respectively.

To solve this, we will follow these steps −

• Define a function dfs(), this will take node, add,

• if node is null, then −

  • return

• if left of node is null and right of node is null and add is non-zero, then −

  • ret := ret + val of node

• dfs(left of node, false)

• dfs(right of node, true)

• From the main method, do the following −

• ret := 0

• dfs(root, true)

• return ret

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class TreeNode{
   public:
   int val;
   TreeNode *left, *right;
   TreeNode(int data){
      val = data;
      left = NULL;
      right = NULL;
   }
};
class Solution {
   public:
   int ret = 0;
   void dfs(TreeNode* node, bool add){
      if(!node)
      return ;
      if(!node−>left && !node->right && add){
         ret += node−>val;
      }
      dfs(node−>left, false);
      dfs(node−>right, true);
   }
   int solve(TreeNode* root) {
      ret = 0;
      dfs(root, true);
      return ret;
   }
};
main(){
   Solution ob;
   TreeNode *root = new TreeNode(3);
   root−>left = new TreeNode(9);
   root−>right = new TreeNode(10);
   root−>left−>left = new TreeNode(15);
   root−>left−>right = new TreeNode(7);
   cout << (ob.solve(root));
}

Input

TreeNode *root = new TreeNode(3);
root−>left = new TreeNode(9);
root−>right = new TreeNode(10);
root−>left−>left = new TreeNode(15);
root−>left−>right = new TreeNode(7);

Output

17