LeetCode #437 - Path Sum III

最新推荐文章于 2022-08-09 08:26:34 发布

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本文介绍了一种算法，用于解决给定二叉树中路径和等于特定值的问题。通过递归方法，不仅考虑了从根节点开始的路径，还考虑了所有可能的路径组合。

题目描述：

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /  \
    5   -3
   / \    \
  3   2   11
 / \   \
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

给定一个二叉树和一个sum值，求树中有多少条路径满足路径和等于sum，路径不一定经过根结点，但一定是从上到下，由父节点到子节点。由于不一定经过根结点，所以分经过根结点或者不经过根结点（即不一定经过左右节点）两种情况计算，还是用递归的方法实现。

class Solution {
public:
    int pathSum(TreeNode* root, int sum) {
        if(root==NULL) return 0;
        return pathSum(root->left,sum)+pathSum(root->right,sum)+path_sum_from_cur(root,sum);
    }
    
    int path_sum_from_cur(TreeNode* cur,int sum)
    {
        if(cur==NULL) return 0;
        int count=0;
        if(cur->val==sum) count++;
        count+=path_sum_from_cur(cur->left,sum-cur->val);
        count+=path_sum_from_cur(cur->right,sum-cur->val);
        return count;
    }
};