刷题437. Path Sum III

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一、题目说明

题目437. Path Sum III，给定一个二叉树和整数sum，计算路径和是sum的数量，其中路径只能是从父节点向下的。难度是Easy！

二、我的解答

这个题目绝对不是Easy！最直观的想法，先判断根节点是否有路径，然后判断左子树，右子树是否有路径。

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为起点，任意节点可作为结束和为sum的个数 
			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、优化措施

网上找了一个dp算法：这个题目类似数组求连续和，将递归算法消除一层递归！

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.

刷题437. Path Sum III

标签：说明   sage   line   sub   roo   刷题   less   ast   solution   

原文地址：https://www.cnblogs.com/siweihz/p/12330038.html