leetCode 113. Path Sum II 二叉樹問題 | Medium

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]
]

思路：

先序遍歷，獲取目標(biāo)序列存到結(jié)果序列中。

代碼如下：

/**
 * 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<vector<int>> result;
        vector<TreeNode *> temp;
        
        DFS(root,result,temp,0,sum);
        return result;
    }
    
    void DFS(TreeNode* root,vector<vector<int>> &result,vector<TreeNode *> &tempNodePtr ,int curTotal , int sum)
    {
        if(!root)
            return;
        curTotal += root->val;
        tempNodePtr.push_back(root);
        vector<int> tempInt;
        if( !root->left && !root->right && curTotal == sum)
        {
            for(int i = 0 ; i < tempNodePtr.size(); i++)
            {
                tempInt.push_back(tempNodePtr[i]->val);
            }
            result.push_back(tempInt);
            tempInt.clear();
        }
        vector<TreeNode *> tempNodePtrLeft(tempNodePtr);
        vector<TreeNode *> tempNodePtrRight(tempNodePtr);
        
        DFS(root->left,result,tempNodePtrLeft,curTotal,sum);
        DFS(root->right,result,tempNodePtrRight,curTotal,sum);
    }
};

2016-08-07 13:52:07