PathSum

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.

Solution

递归题

• 如果p是叶子节点，则若p的值等于sum，return true， 否则return false
• 若p不是叶子节点，则一定存在孩子节点，左孩子或者右孩子满足其中一个和为sum - p->val即可,即 return hasPath(p->left, sum - p->val) || hasPath(p->right, sum - p->right)

Code

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

扩展

见Path Sum II.