LeetCode (Sum Root to Leaf Numbers)
2017-06-26 18:09
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Problem:
Given a binary tree containing digits from
An example is the root-to-leaf path
Find the total sum of all root-to-leaf numbers.
For example,
The root-to-leaf path
The root-to-leaf path
Return the sum = 12 + 13 =
Solution:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int sumNumbers(TreeNode* root) {
return sum(root, 0);
}
int sum(TreeNode* root, int n){
if(!root) return 0;
n = 10 * n + root->val;
if(!root->left && !root->right)
return n;
return sum(root->left, n) + sum(root->right, n);
}
};
Given a binary tree containing digits from
0-9only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path
1->2->3which 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->2represents the number
12.
The root-to-leaf path
1->3represents the number
13.
Return the sum = 12 + 13 =
25.
Solution:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int sumNumbers(TreeNode* root) {
return sum(root, 0);
}
int sum(TreeNode* root, int n){
if(!root) return 0;
n = 10 * n + root->val;
if(!root->left && !root->right)
return n;
return sum(root->left, n) + sum(root->right, n);
}
};
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