Path Sum LeetCode Solution | Easy Approach

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Path Sum 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 Solutions

Time: O(n)
Space: O(h)

C++

class Solution {
public:
bool hasPathSum(TreeNode* root, int sum) {
if (!root)
return false;
if (root->val == sum && !root->left && !root->right)
return true;
return hasPathSum(root->left, sum – root->val) ||
hasPathSum(root->right, sum – root->val);
}
};

Java

 class Solution {
  public boolean 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);
  }
}

Python

class Solution:
  def hasPathSum(self, root: TreeNode, sum: int) -> bool:
    if not root:
      return False
    if root.val == sum and not root.left and not root.right:
      return True
    return self.hasPathSum(root.left, sum - root.val) or \
        self.hasPathSum(root.right, sum - root.val)

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