Path Sum II

最新推荐文章于 2024-10-26 02:26:14 发布

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题目：

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For 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]
]

算法思想：万能的递归，二叉树的不二选择。

vector > pathSum(TreeNode *root, int sum) 
{
    if (root == NULL) return vector>();
    
    if (root->left == NULL && root->right == NULL && sum == root->val)
        return vector>(1, vector(1, root->val));
    
    vector> res_left = pathSum(root->left, sum-root->val);
    vector> res_right = pathSum(root->right, sum-root->val);
    
    for (int i = 0; i < res_left.size(); i++)
        res_left[i].insert(res_left[i].begin(), root->val);
        
    for (int i = 0; i < res_right.size(); i++)
    {
        res_right[i].insert(res_right[i].begin(), root->val);
        res_left.push_back(res_right[i]);
    }
    
    return res_left;
}