# 1022. 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` is `0` or `1`.

``````class Solution {
public:
int sumRootToLeaf(TreeNode* root) {
int res = 0;
helper(root, 0, res);
return res;
}
void helper(TreeNode* node, int cur, int& res) {
if (!node) return;
cur = cur * 2 + node->val;
if (!node->left && !node->right) {
res += cur;
}
helper(node->left, cur, res);
helper(node->right, cur, res);
}
};
``````

``````class Solution {
public:
int sumRootToLeaf(TreeNode* root) {
return helper(root, 0);
}
int helper(TreeNode* node, int cur) {
if (!node) return 0;
cur = cur * 2 + node->val;
return node->left == node->right ? cur : helper(node->left, cur) + helper(node->right, cur);
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1022

Path Sum IV

Path Sum III

Binary Tree Maximum Path Sum

Path Sum II

Path Sum

Minimum Path Sum

https://leetcode.com/problems/sum-of-root-to-leaf-binary-numbers/

https://leetcode.com/problems/sum-of-root-to-leaf-binary-numbers/discuss/270600/Java-Simple-DFS

https://leetcode.com/problems/sum-of-root-to-leaf-binary-numbers/discuss/270025/JavaC%2B%2BPython-Recursive-Solution

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