Path Sum

Q:
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.

A:

-------------------第二遍-------------------

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;
        if(root.left != null && hasPathSum(root.left, sum-root.val))
            return true;
        if(root.right != null && hasPathSum(root.right, sum-root.val))
            return true;
        return false;
    }
} 

----------3 rd pass ---------下面的这个代码是不对的，为什么呢？-----

public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root==null)
            return false;
        return helper(root,sum);
    }
    private boolean helper(TreeNode root, int sum){
        if(root ==null)
            return sum==0;
        return helper(root.left, sum-root.val) || helper(root.right,sum-root.val);
    }
}

Mistakes:

1 :因为，对leaf的理解不对，只有当某个节点没有child的时候，它才是leaf。
而上面的方法，会让不是leaf的节点被当作节点
2:  即使是boolean类型的，要用的时候，也要  先显著的声明下，而不能用它的默认值。
（貌似避免不同平台的默认值不同）
Learned: