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Depth-first-search - Medium Level - Question 1

Depth-first-search - Medium Level - Question 1


Leetcode 113. Path Sum II

Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where each path's sum equals targetSum.

A leaf is a node with no children.

Constraints:

The number of nodes in the tree is in the range [0, 5000].

-1000 <= Node.val <= 1000

-1000 <= targetSum <= 1000


Analysis:

If the root is null, we can stop searching.

Starting with the root, we only have two ways to go: the left child and the right child.

Once we move on to the next step, we need to update the conditions: the sum and the path.

When the node is a leaf, we should check the sum, to see whether it reaches 0 or not (here we use the targetSum - node.val. You can use the sum of all the nodes on the path, but need one more variable).

If the sum is 0, then we find one path, so we can save it in the final answer; if not, we will stop this route and continue the search with other routes.

Since we switch from on path to the other (from the left child to the right child, we just need to pop out the last element in the path, so a path with reference (or in macro) can be used.


See the code below:


/**
 * 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> path;
        dsf(root, sum, res, path);
        return res;
    }
private:
    void dsf(TreeNode* r, int s, vector<vector<int>>& res, vector<int>& ps){
        if(!r) return;
        ps.push_back(r->val);
        if(!r->left && !r->right && s == r->val) res.push_back(ps);
        dsf(r->left, s - r->val, res, ps);
        dsf(r->right, s - r->val, res, ps);
        ps.pop_back();
    }
};


Upper Layer


 

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