Sum Root to Leaf Numbers
描述
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.
For example,
1
/ \
2 3
The root-to-leaf path 1->2
represents the number 12
.
The root-to-leaf path 1->3
represents the number 13
.
Return the sum = 12 + 13 = 25
.
分析
无
代码
- Python
- Java
- C++
// Sum root to leaf numbers
// 时间复杂度O(n),空间复杂度O(logn)
public class Solution {
public int sumNumbers(TreeNode root) {
return dfs(root, 0);
}
private static int dfs(TreeNode root, int sum) {
if (root == null) return 0;
if (root.left == null && root.right == null)
return sum * 10 + root.val;
return dfs(root.left, sum * 10 + root.val) +
dfs(root.right, sum * 10 + root.val);
}
}
// Sum root to leaf numbers
// 时间复杂度O(n),空间复杂度O(logn)
class Solution {
public:
int sumNumbers(TreeNode *root) {
return dfs(root, 0);
}
private:
int dfs(TreeNode *root, int sum) {
if (root == nullptr) return 0;
if (root->left == nullptr && root->right == nullptr)
return sum * 10 + root->val;
return dfs(root->left, sum * 10 + root->val) +
dfs(root->right, sum * 10 + root->val);
}
};
# Sum root to leaf numbers
# 时间复杂度O(n),空间复杂度O(logn)
class Solution:
def sumNumbers(self, root):
return self.dfs(root, 0)
def dfs(self, root, sum):
if not root:
return 0
if not root.left and not root.right:
return sum * 10 + root.val
return self.dfs(root.left, sum * 10 + root.val) + \
self.dfs(root.right, sum * 10 + root.val)