Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

A leaf is a node with no children.

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Example 1:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output: true
Explanation: The root-to-leaf path with the target sum is shown.


Example 2:

Input: root = [1,2,3], targetSum = 5
Output: false
Explanation: There two root-to-leaf paths in the tree:
(1 --&gt; 2): The sum is 3.
(1 --&gt; 3): The sum is 4.
There is no root-to-leaf path with sum = 5.


Example 3:

Input: root = [], targetSum = 0
Output: false
Explanation: Since the tree is empty, there are no root-to-leaf paths.


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Constraints:


	The number of nodes in the tree is in the range [0, 5000].
	-1000 &lt;= Node.val &lt;= 1000
	-1000 &lt;= targetSum &lt;= 1000


class Solution(object):
    def hasPathSum(self, root, targetSum):
        """
        :type root: TreeNode
        :type targetSum: int
        :rtype: bool
        """
        if not root: return False
        targetSum -= root.val
        if not root.left and not root.right:
            return not targetSum
        return self.hasPathSum(root.left, targetSum) or self.hasPathSum(root.right, targetSum)
class Solution {
    public boolean hasPathSum(TreeNode root, int targetSum) {
       if(root == null) return false;
      
      if(root.left == null && root.right == null) return root.val == targetSum;
      
      return hasPathSum(root.right, targetSum - root.val) || hasPathSum(root.left, targetSum - root.val);
    }
}
class Solution {
public:
    bool ans=false;
      int sum=0;
    void recur(TreeNode* root, int target){
        if(root==NULL)return;
        
            sum+=root->val;
            recur(root->left,target);
            recur(root->right,target);
           if(root->left==NULL && root->right==NULL&&sum == target){ // !!Check only if it is a leaf node....
             ans = true;
             return;
            }
            sum-=root->val; //backtracking
            return;
        
    }
    bool hasPathSum(TreeNode* root, int targetSum) {
        recur(root,targetSum);
        return ans;
    }
};
var hasPathSum = function(root, targetSum) {
	return DFS(root, 0 , targetSum)
};

const DFS = (curr, currentSum, targetSum) =>{

	if(!curr) return false

	currentSum += curr.val;

	if(!curr.left && !curr.right) return currentSum === targetSum;

	return DFS(curr.left, currentSum, targetSum) || DFS(curr.right, currentSum, targetSum)
}
Path Sum