112. 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.

解法 :
          基本上就是用DFS, DFS有兩種。一種是遞迴，另外一種是用stack

遞迴法

    邏輯大概是這樣，遞迴的function中先檢查傳進來的節點的三種條件
1. nullptr
    這不用多說，直接回false

2. 葉子
    如果是的話就在檢查傳進來的值是不是等於此節點的值，是的話就回傳true,不是就是false

3. 不是上面兩種情況
     將此節點的左右子節點帶入遞迴的function然後傳入sum - 此節點的值，取兩者遞迴function的OR值

/**
 * Definition for a binary tree node.
 * 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 == nullptr)
            return false;
        
        return hasPathSumDFS(root, sum);    
    }
    
    bool hasPathSumDFS(TreeNode* node, int sum)
    {
        if (node == nullptr)
        {
            return false;
        }
        
        if (node->left == nullptr && node->right == nullptr)
        {
            if(node->val == sum)
            {
                return true;
            }
            else
            {
                return false;
            }
        }

        return hasPathSumDFS(node->left, sum - node->val) || hasPathSumDFS(node->right, sum - node->val);

    }
};