LeetCode 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.

解法一：直接深度优先算法

/**

 * 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 == NULL) return false;

        if (root->left == NULL && root->right == NULL && root->val == sum) return true;

        return hasPathSum(root->left , sum - root->val) | hasPathSum(root->right , sum-root->val);

    }

};

解法二：标准DFS模板

/**

 * 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 mybool = false;

    void dfs(TreeNode* root , int sum , int currentsum) {

        currentsum += root->val;

        bool isleaf = root->left == NUL
91c1
L && root->right == NULL;

        if (isleaf && currentsum == sum) mybool =  true;

        if (root->left) dfs(root->left , sum , currentsum);

        if (root->right) dfs (root->right, sum, currentsum);

    }

    bool hasPathSum(TreeNode* root, int sum) {

        if (root == NULL) return false;

        dfs(root , sum , 0);

        return mybool;

    }

};