437. Path Sum III

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博客围绕二叉树展开，要找出树中路径和为给定值的路径数量。路径无需从根节点或叶子节点开始或结束，但必须向下。还提到递归加深度优先搜索（dfs）的方法，且找到合适路径统计数量时不要返回，因为后续结点和可能为0。

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

递归+dfs

注意当出现合适路径的时候，统计路径数+1没错，但是不要return，因为后面路径的结点和可能为0。比如说你已经找到一条1->2>3和为6的路径，但是1->2->3->4->-4也是和为6的路径。

/**
 * 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 ans=0;
    
    int pathSum(TreeNode* root, int sum) {
        fun(root,sum);
        return ans;
    }
    
    void fun(TreeNode* root, int sum){
        if(root){
            dfs(root,sum,root->val);
            if(root->left){
                fun(root->left,sum);
            }
            if(root->right){
                fun(root->right,sum);
            }
        }        
    }
    
    void dfs(TreeNode* root,int sum,int nowsum){
        if(!root){
            return;
        }
        // if(nowsum>sum){
        //     return;
        // }
        if(nowsum==sum){
            ans++;
//            return;//不要返回
        }
        if(root->left){
            dfs(root->left,sum,nowsum+root->left->val);
        }
        if(root->right){
            dfs(root->right,sum,nowsum+root->right->val);
        }
    }
};

看了别人的写法，很漂亮

class Solution {
public:
    int pathSum(TreeNode* root, int sum) {
        if (!root) return 0;
        int res = findPath(root, 0, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
        return res;
    }
    
    int findPath(TreeNode* node, int curSum, int sum) {
        if (!node) return 0;
        curSum += node->val;
        return (curSum == sum) + findPath(node->left, curSum, sum) + findPath(node->right, curSum, sum);
    }
};