leetcode-112-path sum I

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

遇见树能用递归就用递归,okay!

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转载自blog.csdn.net/vigorqq/article/details/80962760