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 == NULL && 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;
    }
};