acwing 47. 二叉樹中和爲某一值的路徑

地址 https://www.acwing.com/problem/content/description/45/

輸入一棵二叉樹和一個整數，打印出二叉樹中結點值的和爲輸入整數的全部路徑。

從樹的根結點開始往下一直到葉結點所通過的結點造成一條路徑。

給出二叉樹以下所示，並給出num=22。
      5
     / \
    4   6
   /   / \
  12  13  6
 /  \    / \
9    1  5   1

輸出：[[5,4,12,1],[5,6,6,5]]

 

樹的處理 一半是遞歸

/**
 * 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>> ret;

void findPathInner(TreeNode* root, int sum, vector<int> v)
{
    sum -= root->val;
    v.push_back(root->val);

    if (sum == 0 && root->left == NULL && root->right == NULL) {
        ret.push_back(v);
        return;
    }

    if(sum <0 ) return;


    if (root->left != NULL) {
        findPathInner(root->left, sum, v);
    }
    if (root->right != NULL) {
        findPathInner(root->right, sum, v);
    }

}


vector<vector<int>> findPath(TreeNode* root, int sum) {
    if (root == NULL) return ret;
    if (sum == 0) return ret;

    vector<int> v;
    sum -= root->val;
    v.push_back(root->val);
    if (sum < 0) return ret;
    if (sum == 0) {
        ret.push_back(v);
        return ret;
    }
    if (root->left != NULL) {
        findPathInner(root->left, sum, v);
    }
    if (root->right != NULL) {
        findPathInner(root->right, sum, v);
    }

    return ret;
}


};