113. Path Sum II

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每日一题 2019 - 04 - 26

题目：

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

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

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

Return:

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

解法:

这个题让找出二叉树上等于给定值和的路径结点，整体来说深度优先遍历就可以解决问题，但是需要注意两个地方：

  • 返回的路径一定是从根节点到最低层的路径记录
  • 每一层完成遍历一定要让当前记录的 vector 与 sum_value 删除当前层的记录，避免影响进一步的搜索；

代码：

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/**
 * 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<int> now;
        vector<vector<int>> res ;
        if(root == NULL)
        {
            return res ;
        }
        int total_sum = 0;
        find_sum(res,now,root,total_sum,sum) ;
        return res ;
    }
    void find_sum(vector<vector<int>>& res , vector<int> now , TreeNode *temp, int &total , int sum)
    {
        if(temp == NULL)
        {
            return ;
        }
        total += temp -> val ;
        now.push_back(temp->val);
        if(temp-> left == NULL && temp -> right == NULL)
        {
            if(total == sum)
                res.push_back(now);
            total -= temp->val;
            now.pop_back();
            return ;
        }
        find_sum(res,now,temp->left,total,sum);
        find_sum(res,now,temp->right,total,sum);
        total -= temp -> val ;
        now.pop_back();
        return ; 
    }
};
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