666. Path Sum IV

最新推荐文章于 2024-01-13 06:18:24 发布

无名_1989

最新推荐文章于 2024-01-13 06:18:24 发布

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文章标签： Tree

本文介绍了一种利用三元数表示二叉树的方法，并通过递归算法求解从根节点到叶子节点的所有路径的总和。该方法首先将每个节点映射到哈希表中，再依据满二叉树特性寻找子节点，最终实现路径总和的计算。

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题目：

If the depth of a tree is smaller than 5, then this tree can be represented by a list of three-digits integers.

For each integer in this list:

The hundreds digit represents the depth D of this node, 1 <= D <= 4.
The tens digit represents the position P of this node in the level it belongs to, 1 <= P <= 8. The position is the same as that in a full binary tree.
The units digit represents the value V of this node, 0 <= V <= 9.

Given a list of ascending three-digits integers representing a binary with the depth smaller than 5. You need to return the sum of all paths from the root towards the leaves.

Example 1:

Input: [113, 215, 221]
Output: 12
Explanation: 
The tree that the list represents is:
    3
   / \
  5   1

The path sum is (3 + 5) + (3 + 1) = 12.

Example 2:

Input: [113, 221]
Output: 4
Explanation: 
The tree that the list represents is: 
    3
     \
      1

The path sum is (3 + 1) = 4.

思路：

为了便于查找方便，我们首先将每个节点映射到一个哈希表中，其中key是节点的位置，value是节点的值。由于符合满二叉树的性质，所以我们很容易根据一个父结点的key找到它的两个左右子孩子所对应的节点。这样这个问题就和其它的path sum问题没有区别了。

class Solution {
public:
    int pathSum(vector<int>& nums) {
        unordered_map<int, int> hash;
        for (auto num : nums) {
            int hundreds = num / 100;
            int tens = (num % 100) / 10;
            int units = num % 10;
            hash[num - units] = units;
        }
        int hundreds = nums[0] / 100;
        int tens = (nums[0] % 100) / 10;
        int total_sum = 0;
        pathSum(hundreds, tens, 0, total_sum, hash);
        return total_sum;
    }
private:
    void pathSum(int hundreds, int tens, int sum, int &total_sum,
                 unordered_map<int, int> &hash) {
        sum += hash[hundreds * 100 + tens * 10];
        int left = hash.count((hundreds + 1) * 100 + (2 * tens - 1) * 10);
        int right = hash.count((hundreds + 1) * 100 + (2 * tens) * 10);
        if (left == 0 && right == 0) {      // this node is a leaf
            total_sum += sum;
            return;
        }
        if (left != 0) {
            pathSum(hundreds + 1, 2 * tens - 1, sum, total_sum, hash);
        }
        if (right != 0) {
            pathSum(hundreds + 1, 2 * tens, sum, total_sum, hash);
        }
    }
};