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.

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.

/**
 * 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;
        int now=0;
        return digui(root,sum,now);
    }
    boolean digui(TreeNode root, int sum,int now){
        now+=root.val;
        if(sum==now&&root.left==null&&root.right==null){
            return true;
        }
        
        boolean left=false,right=false;
     
        if(root.left!=null)
            left= digui(root.left,sum,now);
        if(root.right!=null)
            right= digui(root.right,sum,now);
        return left||right;
   
    }
}

/**
 * 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 && sum == root.val) {
            return true;
        }
        return hasPathSum(root.left, sum-root.val) || hasPathSum(root.right, sum-root.val);
    }
}