Path Sum LeetCode Solution – Queslers

Problem – Path Sum LeetCode Solution

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.

leaf is a node with no children.

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 --> 2): The sum is 3.
(1 --> 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.

Constraints:

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

Path Sum LeetCode Solution in Java

public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) return false;
    
        if(root.left == null && root.right == null && sum - root.val == 0) return true;
    
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}

Path Sum LeetCode Solution in Python

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param root, a tree node
    # @param sum, an integer
    # @return a boolean
    # 1:27
    def hasPathSum(self, root, sum):
        if not root:
            return False

        if not root.left and not root.right and root.val == sum:
            return True
        
        sum -= root.val

        return self.hasPathSum(root.left, sum) or self.hasPathSum(root.right, sum)

Path Sum LeetCode Solution in C++

bool hasPathSum(TreeNode *root, int sum) {
        if (root == NULL) return false;
        if (root->val == sum && root->left ==  NULL && root->right == NULL) return true;
        return hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val);
    }
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