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.

递归dfs，如果当前节点是叶子节点就判断和是否等于sum

bool dfs(TreeNode *node, int sum, int curSum){    if (node == NULL)        return false;    if (node->left == NULL && node->right == NULL)        return curSum + node->val == sum;    return dfs(node->left, sum, curSum + node->val) || dfs(node->right, sum, curSum + node->val);}bool hasPathSum(TreeNode *root, int sum){    return dfs(root, sum, 0);}