129. Sum Root to Leaf Numbers
2017-12-06 15:59
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Given a binary tree containing digits from
a number.
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 =
题意是一条路径,由根结点到叶子结点,代表一个数,根是高位,一直这样下去。例如例子里,有12和13这两个数,然后求所有路径数之和。其实题目挺简单的,就是一个深搜,所以用递归的话是最简洁易懂的,代码如下。
Code(LeetCode运行3ms)
/**
* 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 sumNumbers(root, 0);
}
int sumNumbers(TreeNode* root, int sum) {
if (!root) {
return 0;
}
if (!root -> left && !root -> right) {
return sum * 10 + root -> val;
}
return sumNumbers(root -> left, sum * 10 + root -> val) + sumNumbers(root -> right, sum * 10 + root -> val);
}
};
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.
题意是一条路径,由根结点到叶子结点,代表一个数,根是高位,一直这样下去。例如例子里,有12和13这两个数,然后求所有路径数之和。其实题目挺简单的,就是一个深搜,所以用递归的话是最简洁易懂的,代码如下。
Code(LeetCode运行3ms)
/**
* 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 sumNumbers(root, 0);
}
int sumNumbers(TreeNode* root, int sum) {
if (!root) {
return 0;
}
if (!root -> left && !root -> right) {
return sum * 10 + root -> val;
}
return sumNumbers(root -> left, sum * 10 + root -> val) + sumNumbers(root -> right, sum * 10 + root -> val);
}
};
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