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.

DFS模板题。注意节点数值可能为负数，常用的sum>target的剪枝不能用了，老老实实搜吧。

要求里面是从根到【叶子】节点，所以不满足此条件的和等于sum也不能算。

public boolean hasPathSum(TreeNode root, int sum)
{
if(root==null)
return false;

return dfs(root, 0, sum);
}

private boolean dfs(TreeNode t,int sum,int target)
{
int val=sum+t.val;

if(val==target&&t.left==null&&t.right==null)
return true;
if(t.left!=null)
if(dfs(t.left, val, target))
return true;
if(t.right!=null)
if(dfs(t.right, val, target))
return true;
return false;
}