113. Path Sum II

problem descripition

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]
]

algorithm thought

和之前的path sum一个意思，只是这里需要得到路径上的所有值。这就只能使用回溯法解决问题了。

code

/**
 * 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<vector<int>> res;
        vector<int> tmp;
        helper(res,tmp,sum,root);
        return res;
    }
    void helper(vector<vector<int>>& res,vector<int>&tmp,int sum,TreeNode* root){
        if(root==NULL){
            return;
        }
        if(root->left==NULL&&root->right==NULL){
            if(sum==root->val){
                tmp.push_back(root->val);
                res.push_back(tmp);
                tmp.pop_back();
            }
            return;
        }
        tmp.push_back(root->val);
        helper(res,tmp,sum-root->val,root->left);
        helper(res,tmp,sum-root->val,root->right);
        tmp.pop_back();
        return;
    }
};

algorithm analysis

时间复杂度O(n),对于每个节点都只访问一次。