LeetCode 666. Path Sum IV

原题链接在这里：https://leetcode.com/problems/path-sum-iv/

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:

1. The hundreds digit represents the depth D of this node, 1 <= D <= 4.
2. 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.
3. 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.

题解：

每个数字的前两位就决定了node所在的level 和 position. 

那么这个node的left child 所在位置是 level + 1, 2*position-1. right child 所在位置是 level + 1, 2*position.

把所有node的位置都存起来, 当走到一个node, 它的left 和 right child 都没有记录的时候就说明走到了叶子节点, 此时记录的path sum可以累积到结果中.

Time Complexity: O(n). n是tree 的node数目.

Space: O(n).

AC Java:

class Solution {
    int sum = 0;
    
    public int pathSum(int[] nums) {
        if(nums == null || nums.length == 0){
            return sum;
        }
        
        HashMap<Integer, Integer> hm = new HashMap<>();
        for(int num : nums){
            hm.put(num / 10, num % 10);
        }
        
        dfs(nums[0] / 10, hm, 0);
        return sum;
    }
    
    private void dfs(int rootKey, Map<Integer, Integer> hm, int cur){
        if(!hm.containsKey(rootKey)){
            return;
        }
        
        cur += hm.get(rootKey);
        
        int level = rootKey / 10;
        int pos = rootKey % 10;
        int leftKey = (level + 1) * 10 + 2 * pos - 1;
        int rightKey = leftKey + 1;
        
        if(!hm.containsKey(leftKey) && !hm.containsKey(rightKey)){
            sum += cur;
            return;
        }
        
        dfs(leftKey, hm, cur);
        dfs(rightKey, hm, cur);
    }
}

类似Path Sum III.