Leetcode-113. Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \    / \
7    2  5   1

Return:

[
   [5,4,11,2],
   [5,8,4,5]
]

/**
 * 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:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int>> paths;
        vector<int> path;
        findpath(root,sum,path,paths);
        return paths;
        
    }
    
    void findpath(TreeNode* root, int sum, vector<int> &path, vector<vector<int>> &paths){
        if(root==NULL){
            return;
        }
        path.push_back(root->val);
        if(root->val==sum&&root->left==NULL&&root->right==NULL){
            paths.push_back(path);
        }
        findpath(root->left,sum-root->val,path,paths);
        findpath(root->right,sum-root->val,path,paths);
        path.pop_back();
    }
};

之前和别人讨论递归函数返回值问题，这里就是一个不需要返回值的递归函数，但其也非常有存在的必要，仅仅是用递归去遍历二叉树。而且在遍历到最后的一个节点之前，我们可以知道这个算法是不可能结束的，所以不用再中途退出，遍历就完事了。