1. 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.

For 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.

Solution.

if you reach to a node which is empty, which means there is no such path, when you reach to a leaf node, check wether it is a valid path. otherwise continue to next level.

/**
 * 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);

    }
}

/**
 * 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;
        return has(root, 0, sum);
    }

    boolean has(TreeNode root, int cur, int target){
        if(root.left == null && root.right == null){
            return target == cur+root.val;
        }
        cur += root.val;

        boolean res = false;
        if(root.left != null){
            res = has(root.left, cur, target);
        }
        if(res ) return true;

        if(root.right != null){
            res = has(root.right, cur,target);
        }

        return res;

    }
}