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.
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool hasPathSum(TreeNode* root, int sum) { 13 if(root == NULL) 14 return false; 15 if(root->left == NULL && root->right == NULL && sum == root->val) 16 return true; 17 if(hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val)) 18 return true; 19 else 20 return false; 21 } 22 };
1 bool hasPathSum(TreeNode* root, int sum){ 2 if(root == NULL) 3 return false; 4 queue<TreeNode*> q; 5 q.push(root); 6 TreeNode* temp; 7 while(!q.empty()){ 8 int size = q.size(); 9 while(size--){ 10 temp = q.front(); 11 q.pop(); 12 if(temp->left != NULL){ 13 q.push(temp->left); 14 temp->left->val += temp->val; 15 } 16 if(temp->right != NULL){ 17 q.push(temp->right); 18 temp->right->val += temp->val; 19 } 20 if(temp->left == NULL && temp->right == NULL && (temp->val == sum)){ 21 return true; 22 } 23 } 24 } 25 return false; 26 }