leetcode 113. Path Sum II【回溯】

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],
   [5,8,4,5]
]

除了num.erase(num.end());不能用 要写 num.pop_back(); 都很完美

/**
 * 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>> ans;
    void cal(vector<int> num,TreeNode* tmp, int tot, int sum) {
        if(tmp->left == NULL && tmp->right == NULL) {
           // printf("tot=%d\n", tot);
            if(tot == sum) {
                ans.push_back(num);
            }
           // return;
        }
        if (tmp->right) {
            //printf("push:%d\n", tmp->right->val);
            num.push_back(tmp->right->val);
            cal(num, tmp->right, tmp->right->val + tot, sum);
            num.pop_back();
           // printf("size:%d\n",num.size());
        }
        if (tmp->left) {
           // printf("push:%d\n", tmp->left->val);
            num.push_back(tmp->left->val);
           // printf("tot=%d\n",tmp->left->val + tot);
            cal(num, tmp->left, tmp->left->val + tot, sum);
            num.pop_back();
           // printf("size:%d\n",num.size());
        }

    }
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<int> num;
        if(root) {
            num.push_back(root->val);
            cal(num, root, root->val, sum);
        }
        return ans;
    }
};