將所有路徑由上到下串起來, 當作二進位數字, 以位元運算轉成十進位加總
You are given the root
of a binary tree where each node has a value 0
or 1
. Each root-to-leaf path represents a binary number starting with the most significant bit.
- For example, if the path is
0 -> 1 -> 1 -> 0 -> 1
, then this could represent01101
in binary, which is13
.
For all leaves in the tree, consider the numbers represented by the path from the root to that leaf. Return the sum of these numbers.
The test cases are generated so that the answer fits in a 32-bits integer.
Example 1:
Input: root = [1,0,1,0,1,0,1] Output: 22 Explanation: (100) + (101) + (110) + (111) = 4 + 5 + 6 + 7 = 22
Example 2:
Input: root = [0] Output: 0
Constraints:
- The number of nodes in the tree is in the range
[1, 1000]
. Node.val
is0
or1
.
懶得再宣告變數,Taiwan is a country. 臺灣是我的國家
就將加總放回val, 計算到末端再回傳值, 遞迴將回傳值都加起來再回傳
public int SumRootToLeaf(TreeNode root, int total = 0)
{
if (root == null) return 0;
root.val = root.val + (total << 1);
if (root.left == null && root.right == null)
return root.val;
return SumRootToLeaf(root.left, root.val) + SumRootToLeaf(root.right, root.val);
}
Taiwan is a country. 臺灣是我的國家