树 Path Sum

方法一：迭代，BFS。为每个节点保存一个从root到当前节点的sum和，若到了叶子节点，比对该sum和与给定的sum是否相等。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode *root, int sum) {
        if(root == NULL) return false;
        stack<pair<TreeNode*,int>> s;
        s.push(make_pair(root,root->val));
        while(!s.empty()) {
            auto node = s.top().first;
            auto tree_sum = s.top().second;
            s.pop();
            if(node->left == NULL && node->right == NULL && tree_sum == sum) {
                return true;
            }
            if(node->left != NULL) {
                s.push(make_pair(node->left, tree_sum + node->left->val));
            }
            if(node->right != NULL) {
                s.push(make_pair(node->right, tree_sum + node->right->val));
            }
        }
        return false;        
    }
};

方法二：递归

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode *root, int sum) {
        if(root == NULL) return false;
        if(root->left == NULL && root->right == NULL) {
            return root->val == sum;
        }
        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);       
    }
};