[LeetCode] Path Sum

標籤：

This problem has a very easy recursive code as follows.

 1 class Solution { 2 public: 3     bool hasPathSum(TreeNode* root, int sum) { 4         if (!root) return false; 5         if (!(root -> left) && !(root -> right)) 6             return sum == root -> val; 7         return hasPathSum(root -> left, sum - root -> val) || 8                hasPathSum(root -> right, sum - root -> val); 9     }10 };

If you insist on an iterative solution, this link has a clean iterative solution in Python. I rewrite the code in C++ as follows. In fact, it uses pair<TreeNode*, int> to recor the information, which greatly simplifies the code.

 1 class Solution { 2 public: 3     bool hasPathSum(TreeNode* root, int sum) { 4         if (!root) return false; 5         stack<pair<TreeNode*, int> > pairs; 6         pairs.push(make_pair(root, sum)); 7         while (!pairs.empty()) { 8             pair<TreeNode*, int> pr = pairs.top(); 9             pairs.pop();10             TreeNode* node = pr.first;11             int remain = pr.second;12             if (!(node -> left) && !(node -> right) && remain == node -> val)13                 return true;14             if (node -> left)15                 pairs.push(make_pair(node -> left, remain - node -> val));16             if (node -> right)17                 pairs.push(make_pair(node -> right, remain - node -> val));18         }19         return false;20     }21 };

This link gives a longer iterative solution using postorder traversal.

 1 class Solution { 2 public: 3     bool hasPathSum(TreeNode* root, int sum) { 4         TreeNode* pre = NULL; 5         TreeNode* cur = root; 6         stack<TreeNode*> nodes; 7         int s = 0; 8         while (cur || !nodes.empty()) { 9             while (cur) {10                 nodes.push(cur);11                 s += cur -> val;12                 cur = cur -> left;13             }14             cur = nodes.top();15             if (!(cur -> left) && !(cur -> right) && s == sum)16                 return true;17             if (cur -> right && pre != cur -> right)18                 cur = cur -> right;19             else {20                 pre = cur;21                 nodes.pop();22                 s -= cur -> val;23                 cur = NULL;24             }25         }26         return false;27     }28 };

 

[LeetCode] Path Sum