Sum Root to Leaf Numbers问题及解法
2017-08-12 18:08
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问题描述:
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.
示例:
The root-to-leaf path
The root-to-leaf path
Return the sum = 12 + 13 =
问题分析:
典型的根先序遍历问题,到结点root时,和s = s * 10 + root->val,若root是叶子节点,则总和sum = sum + s;
过程详见代码:
/**
* 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) {
int sum = 0;
sumofpath(root, sum, 0);
return sum;
}
void sumofpath(TreeNode* root,int& sum,int s)
{
if (root == NULL) return;
s = s * 10 + root->val;
if (root->left == NULL && root->right == NULL)
sum += s;
sumofpath(root->left, sum, s);
sumofpath(root->right, sum, s);
}
};
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.
示例:
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.
问题分析:
典型的根先序遍历问题,到结点root时,和s = s * 10 + root->val,若root是叶子节点,则总和sum = sum + s;
过程详见代码:
/**
* 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) {
int sum = 0;
sumofpath(root, sum, 0);
return sum;
}
void sumofpath(TreeNode* root,int& sum,int s)
{
if (root == NULL) return;
s = s * 10 + root->val;
if (root->left == NULL && root->right == NULL)
sum += s;
sumofpath(root->left, sum, s);
sumofpath(root->right, sum, s);
}
};
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