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.


A 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 are 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


🔄 Dry Run Step-by-Step (DFS Style):

Call: hasPathSum(5, 22)

  • Node = 5

  • Remaining = 22 - 5 = 17

Now explore both left and right:

▶ Left Subtree of 5: hasPathSum(4, 17)

  • Node = 4

  • Remaining = 17 - 4 = 13

▶ Left Subtree of 4: hasPathSum(11, 13)

  • Node = 11

  • Remaining = 13 - 11 = 2


→ Left Subtree of 11: hasPathSum(7, 2)

  • Node = 7 (leaf)

  • Check: 2 == 7? ❌

  • Return false

→ Right Subtree of 11: hasPathSum(2, 2)

  • Node = 2 (leaf)

  • Check: 2 == 2? 

  • Return true

Now we bubble up:
→ hasPathSum(11, 13) = true
→ hasPathSum(4, 17) = true
→ hasPathSum(5, 22) = true 

 How It "Decides" Path:

  • It tries all paths using recursion.

  • If any of them is valid, it returns true.

  • No path is “chosen” ahead — it's exploration via DFS.



/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int targetSum) {
       
   if (root ==null) return false;
   if(isLeaf(root))
   {
   return  targetSum==root.val;
   }
 int remaianing=targetSum-root.val;
return hasPathSum(root.left,remaianing) || hasPathSum(root.right,remaianing);

    }

    public static boolean isLeaf(TreeNode root)
    {
        return (root.left==null) && (root.right==null);
    }
}

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