113. 路径总和 II

该题为112. 路径总和的升级版

给定一个二叉树和一个目标和，找到所有从根节点到叶子节点路径总和等于给定目标和的路径。

说明: 叶子节点是指没有子节点的节点。

Solution1：

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/**
 * 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<int> path;
        vector<vector<int> > result;
        if(root){
            dfs(root,0,sum,path,result);
        } 
        return result;
    }
    
    void dfs(TreeNode* node,int currentsum, int sum,vector<int> &path,vector<vector<int> > &result){
        currentsum += node->val;

        path.push_back(node->val);
        //当到达叶子结点，并且路径之和与sum相同
        if(node && !node->left && !node->right ){
            if(currentsum == sum)
            result.push_back(path);
            return;
        }
        if(node->left){
            dfs(node->left,currentsum,sum,path,result);
            //回溯，还原状态
            path.pop_back();
        }
        if(node->right){
            dfs(node->right,currentsum,sum,path,result);
            //回溯，还原状态
            path.pop_back();
        }
        
    }
};

Solution2：

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/**
 * 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<int> path;
        vector<vector<int> > result;
        preorder(root,0,sum,path,result);
        return result;
    }
    void preorder(TreeNode* node, int currentsum, int sum, vector<int> &path, vector<vector<int> > &result){
        if(!node) return;
        currentsum += node->val;
        path.push_back(node->val);
        if(!node->left && !node->right && currentsum == sum){
            result.push_back(path);
        }
        preorder(node->left,currentsum,sum,path,result);
        preorder(node->right,currentsum,sum,path,result);
        //该步可有可无，因为只会递归到叶子节点才返回，又每次递归都在栈中保留了原来该值的副本。
        // currentsum -= node->val;
        path.pop_back();
    }
};