【LeetCode】 113. Path Sum II

Description

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.

給你一棵二元樹和一個總和值，找到從樹根到樹葉的路徑中，所有節點加總等於給予的總和值的路徑。

提示：一個樹葉就是沒有小孩的節點。

Example:

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]
]

Solution

• 用DFS跑所有路線，跑到樹根時去判斷總和有沒有相等，有的話就加到答案中。
• 如果確定二元樹的值都是非負數，可以進行剪枝加速，不過這邊我沒有去測試。

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:
    void DFS(vector<vector<int>>& ans, vector<int> path, TreeNode* node, int sum, int target)
    {
        if(!node)
        {
            return;
        }
        
        path.push_back(node->val);
        sum+=node->val;
        
        if(!node->left && !node->right)
        {
            if(sum == target)
                ans.push_back(path);
            return;
        }
        
        DFS(ans,path,node->left,sum,target);
        DFS(ans,path,node->right,sum,target);
    }
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int>> ans;
        DFS(ans,vector<int>(0,0),root,0,sum);
        return ans;
    }
};

tags: LeetCode C++