Path Sum II

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问题

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]]

解答

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution1 {public:    vector<vector<int>> pathSum(TreeNode* root, int sum) {          vector<int> a;         path(root,sum,a);               return res;    }    void path(TreeNode *root,int sum,vector<int> a)    {         if(root)        {       // if(root==NULL) return res;          if(root->left==NULL && root->right==NULL && sum==root->val)          {              a.push_back(root->val);              res.push_back(a);              //a.clear();          }         else          {            a.push_back(root->val);            path(root->left,sum-root->val,a);            path(root->right,sum-root->val,a);          }        }    }public:   vector<vector<int>> res;};class Solution2 {public:    vector<vector<int>> pathSum(TreeNode* root, int sum) {         // vector<int> a;         path(root,sum);               return res;    }    void path(TreeNode *root,int sum)    {         if(root)        {       // if(root==NULL) return res;          if(root->left==NULL && root->right==NULL && sum==root->val)          {              a.push_back(root->val);              res.push_back(a);              return ;              //a.clear();          }         else          {               a.push_back(root->val);                             if(root->left)              {                                   path(root->left,sum-root->val);                a.pop_back();                                 }                            if(root->right)             {                path(root->right,sum-root->val);                a.pop_back();             }          }        }    }public:   vector<vector<int>> res;   vector<int> a;};class Solution {public:    Solution():a(1000){}    vector<vector<int>> pathSum(TreeNode* root, int sum) {          path(root,sum,0);               return res;    }    void path(TreeNode *root,int sum,int n)    {        if(root==NULL) return ;         if(root)        {           a[n]=root->val;           n++;                     if(root->left==NULL && root->right==NULL && sum==root->val)          {              vector<int> temp(a.begin(),a.begin()+n);              res.push_back(temp);              return ;          }         else          {                     path(root->left,sum-root->val,n);                     path(root->right,sum-root->val,n);           }        }    }public:   vector<vector<int>> res;   vector<int> a;};

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