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.
Notice:
- Special case for empty tree and
sum=0 - Four case none leaf node with only one child
/**
* 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 fun(TreeNode* root, int sum){
if(NULL == root){
if(sum == 0) return true;
else return false;
}else{
if(NULL == root->left && NULL == root->right){
if(sum == root->val) return true;
else return false;
}else if(NULL == root->left && NULL != root->right){
return fun(root->right, sum - root->val);
}else if(NULL != root->left && NULL == root->right){
return fun(root->left, sum - root->val);
}else{
return fun(root->left, sum - root->val) || fun(root->right, sum - root->val);
}
}
}
bool hasPathSum(TreeNode* root, int sum) {
if(NULL == root) return false;
else return fun(root, sum);
}
};
