剑指OFFER 二叉树中和为某一值的路径

Posted 2020-11-24 virgildevil

二叉树中和为某一值的路径

path数组时刻记录着当前所行走的路径

只有当遍历到叶子结点的时候进行 求和的判断 看是否能把路径添加到结果数组res中.

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};*/
class Solution {
public:
    int expect;
    vector<vector<int> >res;
    vector<int> path;
    void recur(TreeNode* node)
    {
        if(node == NULL)return ;
        if(node->left == NULL && node->right == NULL)
        {
            path.push_back(node->val);
            int sum = 0;
            for(int i=0;i<path.size();i++)
            {
                sum += path[i];
            }
            if(sum == expect)
            {
                res.push_back(path);
            }
            path.pop_back();
            return ;
        }
        
        
        path.push_back(node->val);

        recur(node->left);
        recur(node->right);
        
        path.pop_back();
    }
    vector<vector<int> > FindPath(TreeNode* root,int expectNumber) {
        expect = expectNumber;
        if(root == NULL)return res;
        recur(root);
        return res;
    }
};