刷題437. Path Sum III

一、題目說明

題目437. Path Sum III，給定一個二叉樹和整數sum，計算路徑和是sum的數量，其中路徑只能是從父節點向下的。難度是Easy！

二、我的解答

這個題目絕對不是Easy！最直觀的想法，先判斷根節點是否有路徑，然後判斷左子樹，右子樹是否有路徑。

class Solution{
	public:
		int pathSum(TreeNode* root,int sum){
			if(root==NULL) return 0;
			return dfs(root,sum) + pathSum(root->left,sum) + pathSum(root->right,sum);
		}
		int dfs(TreeNode* root,int sum){
			//以root爲起點，任意節點可作爲結束和爲sum的個數 
			if(root==NULL) return 0;
			
			sum = sum - root->val; 
			int cur = sum==0? 1: 0;
			
			return cur + dfs(root->left,sum) + dfs(root->right,sum);
		}
};

性能如下：

Runtime: 24 ms, faster than 49.38% of C++ online submissions for Minimum Path Sum.
Memory Usage: 14.6 MB, less than 50.00% of C++ online submissions for Minimum Path Sum.

三、優化措施

網上找了一個dp算法：這個題目類似數組求連續和，將遞歸算法消除一層遞歸！

class Solution{
	public:
		// dp solution 
		int pathSum(TreeNode* root,int sum){
			if(root==NULL) return 0;
			sums.resize(maxDepth(root)+1,0);
			dfs(root,1,sum);
			return count;
		}
		int maxDepth(TreeNode* root){
			if(root==NULL) return 0;
			return max(maxDepth(root->left),maxDepth(root->right))+1;
		}
		void dfs(TreeNode* root,int level,int sum){
			if(root==NULL) return;
			
			sums[level] = sums[level-1] + root->val;
			for(int i=0;i<level;i++){
				if(sums[level]-sums[i]==sum) count++;
			}
			dfs(root->left,level+1,sum);
			dfs(root->right,level+1,sum);
		}
	private:
		int count = 0;
		vector<int> sums;
};

性能如下：

Runtime: 20 ms, faster than 77.01% of C++ online submissions for Path Sum III.
Memory Usage: 14.6 MB, less than 100.00% of C++ online submissions for Path Sum III.