113. Path Sum II 解题记录

题目描述：

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]
]

解题思路：

利用递归遍历搜索，把节点的数存在数组中，路径上的数都加在一起，每一次分支都创建一个新数组，到最后一个节点的时候进行判断，路径数的和若和sum相同就把数组加在二维数组。

代码：

/**
 * 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:
    void select(vector<vector<int>> &ret, int sum, TreeNode* root, vector<int> path, int now){
        if(!root->left && !root->right){
        //到最后一个节点
            if(sum == now+root->val){
            //路径上的数与sum相同
                path.push_back(root->val);
                ret.push_back(path);
            }
        }
        else{
        //不是最后一个节点
            path.push_back(root->val);
            if(root->left)
                select(ret, sum, root->left, path, now+root->val);
            if(root->right)
                select(ret, sum, root->right, path, now+root->val);
        }
    }
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int>> ret;
        if(!root)
            return ret;
        select(ret, sum, root, vector<int>(), 0);
        return ret;
    }
};