[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.

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]
]

解法一：

/**
 * 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) {
        if(!root) return {};
        vector<vector<int>> res;
        vector<int> out;
        pathSum(res, out, root, sum);
        return res;
    }
    
    void pathSum(vector<vector<int>>&res, vector<int>& out, TreeNode* root, int sum){
        if(!root) return;
        
        if(!root->left&&!root->right){
            if(root->val==sum){
                out.push_back(root->val);
                res.push_back(out);
                out.pop_back();
                return;
            }else return;
        }
        
        out.push_back(root->val);
        pathSum(res, out, root->left, sum-root->val);
        out.pop_back();
        
        out.push_back(root->val);
        pathSum(res, out, root->right, sum-root->val);
        out.pop_back();
    }
};

解法二：

更加簡潔點。

/**
 * 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) {
        if(!root) return {};
        vector<vector<int>> res;
        vector<int> out; 
        out.push_back(root->val);
        pathSum(res, out, root, sum);
        return res;
    }
    
    void pathSum(vector<vector<int>>&res, vector<int>& out, TreeNode* root, int sum){
        if(!root->left&&!root->right&&root->val==sum){
            res.push_back(out);
        }
        
        if(root->left){
            out.push_back(root->left->val);
            pathSum(res, out, root->left, sum-root->val);
            out.pop_back();
        }
        
        if(root->right){
            out.push_back(root->right->val);
            pathSum(res, out, root->right, sum-root->val);
            out.pop_back();
        }
    }
};