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;}