解法一:
/**
* 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==NULL)
return false;
if(root->left==NULL&&root->right==NULL)
return sum==root->val;
return hasPathSum(root->left,sum-root->val)||hasPathSum(root->right,sum-root->val);
}
};
解法二:
/**
* 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,int curSum){
if(root==NULL)
return false;
curSum+=root->val;
if(root->left==NULL&&root->right==NULL)
return sum==curSum;
return hasPathSum(root->left,sum,curSum)||hasPathSum(root->right,sum,curSum);
}
bool hasPathSum(TreeNode *root, int sum) {
return hasPathSum(root,sum,0);
}
};