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.

到叶子所以加(left == null && right == null  && 当前值满足条件) 

后序遍历

public:
    bool hasPathSum(TreeNode *root, int sum) {
        if (root == NULL) return false;
        if (root->left == NULL && root->right == NULL && root->val == sum ) return true;
        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
    }
};

 The basic idea is to subtract the value of current node from sum until it reaches a leaf node and the subtraction equals 0, then we know that we got a hit. Otherwise the subtraction at the end could not be 0.