112. Path Sum

class Solution {
    public boolean hasPathSum(TreeNode root, int t){
        if(root==null) return false;    //input : root =[], t=0 => false
        return solve(root, t);
    }

    public static boolean solve(TreeNode root, int t){
        if(root==null) return false;    // if we reach a null node, return false
        if(root.left==null && root.right==null && t==root.val) return true; //at leaf node, with target
        boolean left = solve(root.left, t-root.val);
        boolean right = solve(root.right, t-root.val);
        return left || right;
    }
}

/*
Note : t can become negative and positive since there ar eboth element elements
so, (t<0) return false is incorrect.

So, for a binary tree with n nodes, the function will visit each node once, resulting in a time complexity of O(n), where n is the total number of nodes in the tree.

In the worst case (for an unbalanced tree, such as a skewed tree), the height of the tree will be n, leading to a space complexity of O(n).
In the best case (for a balanced tree), the height of the tree will be log(n), leading to a space complexity of O(log n).
*/