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113. Path Sum II

Posted on Edited on In Disqus:
Symbols count in article: 790 Reading time ≈ 1 mins.

backtracking + dfs O(n)

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/**
 * 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> v;
        dfs(root, sum, v, res);
        return res;
    }

    void dfs(TreeNode *root, int sum, vector<int> &v, vector<vector<int>> &res) {
        if (!root) return;
        v.push_back(root->val);
        if (!root->left && !root->right) {
            if (sum == root->val) {
                res.push_back(v);
            }
        } else {
            dfs(root->left, sum - root->val, v, res);
            dfs(root->right, sum - root->val, v, res);
        }
        v.pop_back();
    }
};