【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

Java Completions:

/**
 * 递归实现
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int targetSum) {
        if (root == null) return false;
        if (root != null && root.left == null && root.right == null && targetSum == root.val) return true;
        int new_target = targetSum - root.val;
        return hasPathSum(root.left, new_target) || hasPathSum(root.right, new_target);
    }
}

/**
 * 非递归实现
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int targetSum) {
        if (root == null) return false;
        Stack<TreeNode> tree = new Stack<>();
        Stack<Integer> val = new Stack<>();
        tree.push(root);
        val.push(root.val);
        while (!tree.isEmpty()) {
            TreeNode pop = tree.pop();
            Integer popVal = val.pop();
            if (pop.left == null && pop.right == null && popVal == targetSum) return true;
            if (pop.left != null) {
                tree.push(pop.left);
                val.push(popVal + pop.left.val);
            }
            if (pop.right != null) {
                tree.push(pop.right);
                val.push(popVal + pop.right.val);
            }
        }
        return false;
    }
}