【Leetcode】Path Sum

題目：

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,

              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.

解題思路：

如果節點爲空，返回false。如果節點不爲空，則在路徑和上加上當前節點的值，並判斷是否是葉子節點，如果是葉子節點則判斷當前的路徑和是否等於條件所給的值。如果不是葉子節點，則遞歸的訪問節點的左兒子和右兒子並對兩者的結果取或。

代碼1：

/**
 * 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) {
         return hasPathSum(root,sum,0);
    }
private:
    bool hasPathSum(TreeNode *root, int sum, int CurrSum){
        if(!root)return false;
        CurrSum+=root->val;
        if((!root->left)&&(!root->right))return CurrSum==sum;
        return hasPathSum(root->left,sum,CurrSum)||hasPathSum(root->right,sum,CurrSum);
    }
};

代碼2：可以不另寫函數，只需在訪問節點時從sum值中減去節點值，並將差值作爲新的sum值輸入到下一層的遞歸函數中即可。

/**
 * 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)return false;
        if((!root->left)&&(!root->right))return root->val==sum;
        return hasPathSum(root->left,sum-root->val)||hasPathSum(root->right,sum-root->val);
    }
};