LeetCode 112. Path Sum

112. Path Sum

My Submissions QuestionEditorial Solution

Total Accepted: 99234 Total Submissions: 317747 Difficulty: Easy

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.

递归方法：
/**
* 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 && sum == root->val)
return true;
if(hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val))
return true;
else
return false;
}
};

非递归法
/**
* 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) {
queue<TreeNode *> q;
if(root == NULL)
return false;
q.push(root);
TreeNode *temp;
while(!q.empty()) {
int size = q.size();
while(size--) {
temp = q.front();
q.pop();
if(temp->left != NULL) {
q.push(temp->left);
temp->left->val += temp->val;
}
if(temp->right != NULL) {
q.push(temp->right);
temp->right->val += temp->val;
}
if(temp->left == NULL && temp->right == NULL && (temp->val == sum || temp->val == sum)) {
return true;
}
}
}
return false;
}
};