Path Sum

112. 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 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

My Solutions:

    //recursive
    // public boolean hasPathSum(TreeNode root, int sum) {
    //     if (root == null) return false;
    //     if (root.left == null && root.right == null) return sum == root.val;
    //     return (hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val));
    // }
    
    //iterative
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) return false;
        Stack<TreeNode> path = new Stack<>();
        Stack<Integer> sub = new Stack<>();
        path.push(root);
        sub.push(root.val);
        while (!path.isEmpty()) {
            TreeNode temp = path.pop();
            int tempVal = sub.pop();
            if (temp.left == null && temp.right == null) {
                if (tempVal == sum) return true;
            } else {
                if (temp.left != null) {
                    path.push(temp.left);
                    sub.push(temp.left.val + tempVal);
                }
                if (temp.right != null) {
                    path.push(temp.right);
                    sub.push(temp.right.val + tempVal);
                }
            }
        }
        return false;
    }