Path Sum II

发布时间：2020-12-14 04:44:36 所属栏目：大数据来源：网络整理

导读：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 , 5 / 4 8 / / 11 13 4 / / 7 2 5 1 Return:[ [

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 = 22,5
     /     4   8
   /   /   11  13  4
 /      / 7    2  5   1
Return:

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

code

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)
    {
        if(!root)
            return {};

        vector<vector<int>> res;
        vector<int> tmp;
        pathSumCore(root,sum,tmp,res);
        return res;
    }
private:
    void pathSumCore(const TreeNode* node,int sum,vector<int> &tmp,vector<vector<int>> &res)
    {
        tmp.emplace_back(node->val);

        if(node->left==nullptr&&node->right==nullptr)
        {
            if(sum-node->val==0)
                res.emplace_back(tmp);
            return;
        }

        if(node->left)
        {
            pathSumCore(node->left,sum-node->val,res);
            tmp.pop_back();
        }
        if(node->right)
        {
            pathSumCore(node->right,res);
            tmp.pop_back();
        }
    }
};

（编辑：李大同）

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