Path Sum

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.

Note: A leaf is a node with no children.

Example:

Given the below binary tree andsum = 22,

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

return true, as there exist a root-to-leaf path5->4->11->2which sum is 22.

Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        return sum(root, 0, sum);
    }

    private boolean sum(TreeNode node, int current, int sum) {
        if (node == null) {
            return false;
        }
        int val = current + node.val;
        if (node.left == null && node.right == null) {
            if (val == sum) {
                return true;
            }
            return false;
        }
        if (sum(node.left, val, sum)) {
            return true;
        }
        if (sum(node.right, val, sum)) {
            return true;
        }
        return false;
    }
}

Other recursive solution.

public boolean hasPathSum(TreeNode root, int sum) {
    if (root == null) return false;
    if (root.left == null && root.right == null && root.val == sum) return true;
    return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}

Iterative solution.

public boolean hasPathSum(TreeNode root, int sum) {
    // iteration method
    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;
}