Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

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2
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7
      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

Recursive O(n) time solution

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

Iterative O(n) time O(n) space solution

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) return false;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode prev = null, cur = root;
        int sumTree = 0;
        while (cur != null || !stack.isEmpty()){
            while (cur != null) {
                stack.push(cur);
                sumTree += cur.val;
                cur = cur.left;
            }
            cur = stack.peek();
            if (cur.left == null && cur.right == null && sumTree == sum) {
                return true;
            }
            if (cur.right != null && prev != cur.right) {
                cur = cur.right;
            } else {
                prev = cur;
                stack.pop();
                sumTree -= cur.val;
                cur = null;
            }
        }
        return false;
    }
}