80：Path Sum

题目: 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,

该题目只能采取自顶向下的解法，但该解法有递归和使用栈这两种版本

解析1：递归代码如下：

// 递归版，时间复杂度 O(n)，空间复杂度 O(logn)// 自顶向下法class Solution {public:        bool hasPathSum(TreeNode* root, int sum) {                if (!root) return false;                int res = sum - root -> val;                if (!root -> left && !root -> right)                        return res == 0;                return hasPathSum(root -> left, res) || hasPathSum(root -> right, res);        }};

解析2：迭代代码如下：

// 迭代法，时间复杂度O(n)，空间复杂度 O(log n)// 自顶向下法class Solution {public:        bool hasPathSum(TreeNode* root, int sum) {                if (!root) return false;                stack<pair<TreeNode*, int>> s;                s.push(make_pair(root, sum - root -> val));                while (!s.empty()) {                        auto p = s.top().first;                        sum = s.top().second;                        s.pop();                        if (!p -> left && !p -> right && sum == 0)                                return true;                        if (p -> left)                                s.push_back(make_pair(p -> left, sum - p -> left -> val));                        if (p -> right)                                s.push_back(make_pair(p -> right, sum - p -> right -> val));                }                return false;        }}