【LeetCode】113. Path Sum II 基于Java和C++的解法及分析

Given a binary tree and a sum, find all root-to-leaf paths where each path‘s sum equals the given sum.

              5
             /             4   8
           /   /           11  13  4
         /  \    /         7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

/**
 * 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;//存放满足要求的路径
        vector<int> cur;//当前路径
        if(root==NULL)return result;
        
        DFS(result,cur,root,sum);//深度搜索
        return result;
    }
    
    void DFS(vector<vector<int>> &result,vector<int> &cur,TreeNode* root,int sum)
    {
        if(root==NULL)return;
        if(root->left==NULL&&root->right==NULL)//到达叶子节点，形成合法路径，这时候再对路径进行验证
        {
            cur.push_back(root->val);//叶子节点纳入当前路径
            if(sumOfPath(cur)==sum)//检验路径是否满足约束
            result.push_back(cur);
            
            cur.pop_back();//弹出当前路径末端数据并返回
            return;
        }
        else
        {
            cur.push_back(root->val);//中序遍历的构架
            DFS(result,cur,root->left,sum);
            DFS(result,cur,root->right,sum);
            cur.pop_back();//某一根节点的子节点搜索完毕，末端数据弹出，返回上一层
        }
    
    }
    
    int sumOfPath(vector<int> &dataSet)//求取路径和
    {
        int sum=0;
        for(int i=0;i<dataSet.size();i++)
        {
            sum+=dataSet[i];
        }
        return sum;
    }
};

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    
    public List<List<Integer>> pathSum(TreeNode root, int sum) 
    {
	   List<List<Integer>> result=new ArrayList<List<Integer>>();//List接口变量用的ArrayList类对象实例化
	   List<Integer> temp=new ArrayList<Integer>();
	   
	   if(root==null)return result;
	   DFS(result,temp,root,sum);
	   return result;
    }
    
    void DFS(List<List<Integer>> result,List<Integer> temp,TreeNode root,int sum)
    {
	   if(root==null)return;
	   if(root.left==null&&root.right==null)
	   {
		   temp.add(root.val);
		   if(sumOfPath(temp)==sum)
			   result.add(new ArrayList(temp));//一定要注意
		   temp.remove(temp.size()-1);
		   return;
	   }
	   else
	   {
		   temp.add(root.val);
		   DFS(result,temp,root.left,sum);
		   DFS(result,temp,root.right,sum);
		   temp.remove(temp.size()-1);
	   }
   }
   
   int sumOfPath(List<Integer> temp)
   {
	   int sum=0;
	   for(int i=0;i<temp.size();i++)
	   {
		   sum+=temp.get(i);
	   }
	   return sum;
   }
}