README_EN.md

666. Path Sum IV

中文文档

Description

If the depth of a tree is smaller than 5, then this tree can be represented by an array of three-digit integers. For each integer in this array:

• The hundreds digit represents the depth d of this node where 1 <= d <= 4.
• The tens digit represents the position p of this node in the level it belongs to where 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 where 0 <= v <= 9.

Given an array of ascending three-digit integers nums representing a binary tree with a depth smaller than 5, 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.

Solutions

DFS.

Python3

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

Java

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);
    }
}

C++

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);
    }
};

Go

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
}

...