Path Sum II深度优先找路径

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

For 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]
]

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Tree Depth-first Search

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    vector<vector<int> > paths;
public:
    void dfs(TreeNode *node,int sum,int csum,vector<int> onePath){    //a能为引用
        if(node==NULL)
            return;
        if(node->left==NULL && node->right==NULL){
            if(node->val+csum==sum){
                onePath.push_back(node->val);
                paths.push_back(onePath);
            }
            return;
        }
        onePath.push_back(node->val);
        dfs(node->left,sum,csum+node->val,onePath);
        dfs(node->right,sum,csum+node->val,onePath);
    }

    vector<vector<int> > pathSum(TreeNode *root, int sum) {
        paths.clear();
        vector<int> onePath;
        dfs(root,sum,,onePath);
        return paths;
    }
};