112 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 and sum = 22,

      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.

DFS ()

/**
 * 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) {
        if (root == null) {
            return false;
        }
        //preorder traversal
        //decrease the sum
        if (root.left == null && root.right == null) {
            return root.val == sum;
        }
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}

/**
 * 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) {
       if (root == null) return false;
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            if (cur.left == null && cur.right == null) {
                if (cur.val == sum) {
                    return true;
                }
            }
            if (cur.right != null) {
                cur.right.val += cur.val;
                stack.push(cur.right);

            }
            if (cur.left != null) {
                cur.left.val += cur.val;
                stack.push(cur.left);

            }
        }
        return false;
    }
}