Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path 1->2->3 which represents the number 123.
Find the total sum of all root-to-leaf numbers.
Note: A leaf is a node with no children.
Example:
Input: [1,2,3]
1
/ \
2 3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.
Example 2:
Input: [4,9,0,5,1]
4
/ \
9 0
/ \
5 1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.
Related Topics:
Tree, Depth-first Search
Similar Questions:
// OJ: https://leetcode.com/problems/sum-root-to-leaf-numbers/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
int ans = 0;
void dfs(TreeNode *root, int sum) {
if (!root) return;
sum = sum * 10 + root->val;
if (!root->left && !root->right) ans += sum;
dfs(root->left, sum);
dfs(root->right, sum);
}
public:
int sumNumbers(TreeNode* root) {
dfs(root, 0);
return ans;
}
};// OJ: https://leetcode.com/problems/sum-root-to-leaf-numbers/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
public:
int sumNumbers(TreeNode* root) {
if (!root) return 0;
stack<pair<TreeNode*, int>> s;
s.emplace(root, 0);
int ans = 0;
while (s.size()) {
auto p = s.top();
root = p.first;
int sum = p.second * 10 + root->val;
s.pop();
if (!root->left && !root->right) ans += sum;
if (root->right) s.emplace(root->right, sum);
if (root->left) s.emplace(root->left, sum);
}
return ans;
}
};