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666. Path Sum IV πŸ”’

Description

If the depth of a tree is smaller than 5, then this tree can be represented by an array of three-digit integers. You are given an ascending array nums consisting of three-digit integers representing a binary tree with a depth smaller than 5, where for each integer:

  • The hundreds digit represents the depth d of this node, where 1 <= d <= 4.
  • The tens digit represents the position p of this node within its level, where 1 <= p <= 8, corresponding to its position in a full binary tree.
  • The units digit represents the value v of this node, where 0 <= v <= 9.

Return the sum of all paths from the root towards the leaves.

It is guaranteed that the given array represents a valid connected binary tree.

 

Example 1:

Input: nums = [113,215,221]

Output: 12

Explanation:

The tree that the list represents is shown.
The path sum is (3 + 5) + (3 + 1) = 12.

Example 2:

Input: nums = [113,221]

Output: 4

Explanation:

The tree that the list represents is shown. 
The path sum is (3 + 1) = 4.

 

Constraints:

  • 1 <= nums.length <= 15
  • 110 <= nums[i] <= 489
  • nums represents a valid binary tree with depth less than 5.
  • nums is sorted in ascending order.

Solutions

Solution 1

Thinking

Nodes are encoded as depth, position, and value. Building an explicit tree is unnecessary.

Map \(depth\times 10+pos\) to the value. DFS from \(11\) using the child-index formula; when both children are missing, add the path sum.

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class Solution:
    def pathSum(self, nums: List[int]) -> int:
        def dfs(node, t):
            if node not in mp:
                return
            t += mp[node]
            d, p = divmod(node, 10)
            l = (d + 1) * 10 + (p * 2) - 1
            r = l + 1
            nonlocal ans
            if l not in mp and r not in mp:
                ans += t
                return
            dfs(l, t)
            dfs(r, t)

        ans = 0
        mp = {num // 10: num % 10 for num in nums}
        dfs(11, 0)
        return ans

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class Solution {
    private int ans;
    private Map<Integer, Integer> mp;

    public int pathSum(int[] nums) {
        ans = 0;
        mp = new HashMap<>(nums.length);
        for (int num : nums) {
            mp.put(num / 10, num % 10);
        }
        dfs(11, 0);
        return ans;
    }

    private void dfs(int node, int t) {
        if (!mp.containsKey(node)) {
            return;
        }
        t += mp.get(node);
        int d = node / 10, p = node % 10;
        int l = (d + 1) * 10 + (p * 2) - 1;
        int r = l + 1;
        if (!mp.containsKey(l) && !mp.containsKey(r)) {
            ans += t;
            return;
        }
        dfs(l, t);
        dfs(r, t);
    }
}

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class Solution {
public:
    int ans;
    unordered_map<int, int> mp;

    int pathSum(vector<int>& nums) {
        ans = 0;
        mp.clear();
        for (int num : nums) mp[num / 10] = num % 10;
        dfs(11, 0);
        return ans;
    }

    void dfs(int node, int t) {
        if (!mp.count(node)) return;
        t += mp[node];
        int d = node / 10, p = node % 10;
        int l = (d + 1) * 10 + (p * 2) - 1;
        int r = l + 1;
        if (!mp.count(l) && !mp.count(r)) {
            ans += t;
            return;
        }
        dfs(l, t);
        dfs(r, t);
    }
};

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func pathSum(nums []int) int {
    ans := 0
    mp := make(map[int]int)
    for _, num := range nums {
        mp[num/10] = num % 10
    }
    var dfs func(node, t int)
    dfs = func(node, t int) {
        if v, ok := mp[node]; ok {
            t += v
            d, p := node/10, node%10
            l := (d+1)*10 + (p * 2) - 1
            r := l + 1
            if _, ok1 := mp[l]; !ok1 {
                if _, ok2 := mp[r]; !ok2 {
                    ans += t
                    return
                }
            }
            dfs(l, t)
            dfs(r, t)
        }
    }
    dfs(11, 0)
    return ans
}

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