113. Path Sum II [LeetCode]

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113. Path Sum II

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]
]

深度优先搜索，如果没有找到，返回时要记得删除插入的节点，重新搜索另一条路径，再把另一条路径push_back进去，比较是不是结果，如果还不是还是要清理出来再返回到上一节点，继续往下深搜。

class Solution {

public:

    vector <vector < int >> pathSum(TreeNode * root, int sum) {

        vector <vector < int >>result;

        vector < int >cur;

         pathSum(root, sum, cur, result);

         return result;

    }

     void pathSum(TreeNode * root, int sum, vector < int >&cur, vector <vector < int >>&result) {

         if (root == nullptr) return;

        cur. push_back(root-> val);

         if (root-> left == nullptr &&root-> right == nullptr &&sum == root-> val)

            result. push_back(cur);

         pathSum(root-> left, sum - root-> val, cur, result);

         pathSum(root-> right, sum - root-> val, cur, result);

        cur. pop_back();

    }

};