Leetcode 0112. 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.

Easy

Description

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

• The root-to-leaf path with the target sum is shown.

Example 2:

Input: root = [1,2,3], targetSum = 5
Output: false

• 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

• 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

Solution

/**
 * 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;
        Stack<TreeNode> stack = new Stack<>();
        Stack<Integer> sum = new Stack<>();
        stack.push(root);
        sum.push(root.val);
        while (!stack.isEmpty()) {
            TreeNode curr = stack.pop();
            int currValue = sum.pop();

            if (curr.left == null && curr.right == null) {
                if (currValue == targetSum)
                    return true;
            }
            else {
                if (curr.left != null) {
                    stack.push(curr.left);
                    sum.push(curr.left.val + currValue);
                }
                if (curr.right != null) {
                    stack.push(curr.right);
                    sum.push(curr.right.val + currValue);
                }
            }
        }
        return false;
    }
}