666. Path Sum IV

方法

we can use num / 10 = 10 * depth + pos as a unique identifier for that node. The left child of such a node would have identifier 10 * (depth + 1) + 2 * pos - 1, and the right child would be one greater.

//           xyv
//     /            \
// (x+1)(2y-1)v   (x+1)(2y)v 

class Solution {
    int sum = 0;
    // treenode_index to treenode_val
    Map<Integer, Integer> map = new HashMap<>();
    public int pathSum(int[] nums) {
        for (int num : nums) {
            int val = num % 10;
            int level = num / 100, pos = num / 10 % 10;
            map.put(level * 10 + pos, val);
        }
        int root = nums[0] / 100 * 10 + nums[0] / 10 % 10;
        dfs(root, nums[0] % 10); // count in current node to sum
        return sum;
    }

    public void dfs(int node, int curSum) {
        int level = node / 10, pos = node % 10;
        int left = (level + 1) * 10 + (2 * pos - 1);
        int right = (level + 1) * 10 + (2 * pos);
        if (!map.containsKey(left) && !map.containsKey(right)) {
            sum += curSum;
            return;
        }
        if (map.containsKey(left)) {
            dfs(left, curSum + map.get(left));
        }
        if (map.containsKey(right)) {
            dfs(right, curSum + map.get(right));
        }
    }
}