Tuesday, August 29, 2017

LeetCode 666 - Path Sum IV


http://www.vkadoo.cn/B03735571DC135B35FE7C2ED39549238.AHtml
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.
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.
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.

https://discuss.leetcode.com/topic/101111/java-solution-represent-tree-using-hashmap
How do we solve problem like this if we were given a normal tree? Yes, traverse it, keep a root to leaf running sum. If we see a leaf node (node.left == null && node.right == null), we add the running sum to the final result.
Now each tree node is represented by a number. 1st digits is the level, 2nd is the position in that level (note that it starts from 1 instead of 0). 3rd digit is the value. We need to find a way to traverse this tree and get the sum.
The idea is, we can form a tree using a HashMap. The key is first two digits which marks the position of a node in the tree. The value is value of that node. Thus, we can easily find a node's left and right children using math.
Formula: For node xy? its left child is (x+1)(y*2-1)? and right child is (x+1)(y*2)?
Given above HashMap and formula, we can traverse the tree
    int sum = 0;
    Map<Integer, Integer> tree = new HashMap<>();
    
    public int pathSum(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        
        for (int num : nums) {
            int key = num / 10;
            int value = num % 10;
            tree.put(key, value);
        }
        
        traverse(nums[0] / 10, 0);
        
        return sum;
    }
    
    private void traverse(int root, int preSum) {
        int level = root / 10;
        int pos = root % 10;
        int left = (level + 1) * 10 + pos * 2 - 1;
        int right = (level + 1) * 10 + pos * 2;
        
        int curSum = preSum + tree.get(root);
        
        if (!tree.containsKey(left) && !tree.containsKey(right)) {
            sum += curSum;
            return;
        }
        
        if (tree.containsKey(left)) traverse(left, curSum);
        if (tree.containsKey(right)) traverse(right, curSum);
    }
https://discuss.leetcode.com/topic/101116/c-java-clean-code
         0
     0       1
  0   1     2   3
0 1  2 3   4 5  6 7
Regardless whether these nodes exist:
the position of left child is always parent_pos * 2;
the position of right child is alwaysparent_pos * 2 + 1;
the position of parent is always child_pos / 2;



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