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