Path-Sum

第63天。

赶算法实验，再水一题。

今天的题目是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.

比较简单，但是有一些坑点。

它要求一定要到left.
然后空节点不能当成0.

然后是代码：

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

其实dicuss中的更精炼一点：

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