437. Path Sum III
You are given a binary tree in which each node contains an integer value.
Find the number of paths that sum to a given value.
The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).
The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.
Example:
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8
10
/ \
5 -3
/ \ \
3 2 11
/ \ \
3 -2 1
Return 3. The paths that sum to 8 are:
1. 5 -> 3
2. 5 -> 2 -> 1
3. -3 -> 11题目链接：https://leetcode-cn.com/problems/path-sum-iii/
思路
本题的sum是从任意位置开始，任意位置结束的。
注意一点：由于树中有负值元素，路径可以覆盖。中途找到解不能直接终止递归，因为可能往下走还能满足条件。
法一：递归
思路一：每次考虑当前节点为结束节点
时间复杂度O(n)，空间复杂度不考虑递归是O(logn)。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int pathSum(TreeNode* root, int sum) {
if(!root) return 0;
vector<int> rec;
return path(root, sum, rec);
}
int path(TreeNode* root, int sum, vector<int> rec){
if(!root) return 0;
int res = 0;
for(int i=0;i<rec.size();++i){
rec[i] += root->val;
if(rec[i]==sum) ++res;
}
if(root->val == sum) ++res;
rec.push_back(root->val);
return res + path(root->left, sum, rec) + path(root->right, sum, rec);
}
};编程笔记：递归中传入引用变量，效果是所有层递归函数公用同一个变量，中途被改变，层都能收到；不是引用则只要本函数内被改变，和传递给下层的被改变。
思路二：每次考虑当前节点为开始节点
时间复杂度O(nlgn)，空间复杂度不考虑递归是O(1)。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int pathSum(TreeNode* root, int sum) {
if(!root) return 0;
return path(root, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
}
int path(TreeNode* root, int sum){
if(!root) return 0;
int l = path(root->left, sum-root->val);
int r = path(root->right, sum-root->val);
return (sum==root->val)?1+l+r:l+r;
}
};法二：非递归
暂时不写
