Sum of Root To Leaf Binary Numbers
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 represent 01101
in binary, which is 13
.
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 answer is guaranteed to fit 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
Example 3:
Input: root = [1]
Output: 1
Example 4:
Input: root = [1,1]
Output: 3
Constraints:
The number of nodes in the tree is in the range
[1, 1000]
.Node.val
is0
or1
.
Solutions
🧠 Cpp
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution
{
//list of binary represented
static list<list<int>> getAllNumbers(TreeNode* root)
{
if(!root)
return {};
//if leaf
if(!root->left && !root->right)
return list<list<int>>{{root->val}};
list<list<int>> res;
auto process_leaf = [&](TreeNode* leaf)
{
if(leaf)
for(auto num : getAllNumbers(leaf))
{
num.push_front(root->val);
res.push_back(num);
}
};
process_leaf(root->left);
process_leaf(root->right);
return res;
}
public:
int sumRootToLeaf(TreeNode* root)
{
int res = 0;
for(auto num : getAllNumbers(root))
{
string binary_string;
for(int bit : num)
binary_string.push_back(to_string(bit)[0]);
res += std::stoi(binary_string, nullptr, 2);
}
return res;
}
};
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