leetcode Sum Root to Leaf Numbers
2013-07-09 20:42
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题意:给定一个二叉树,从跟节点到每一个叶子节点产生一个整数,将产生的所有整数相加,返回最终的和。
解题思路:这道题还是利用dfs算法解决,通过递归遍历得到最终的结果。
代码实现:
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int sumNumbers(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int sum = 0, temp = 0;
if(root == NULL)
{
return 0;
}
search(root, sum, temp);
return sum;
}
void search(TreeNode *p, int &sum, int temp)
{
if(p->left == NULL && p->right == NULL)
{
temp = temp * 10 + p->val;
sum += temp;
return;
}
temp = temp * 10 + p->val;
if(p->left != NULL)
{
search(p->left, sum, temp);
}
if(p->right != NULL)
{
search(p->right, sum, temp);
}
}
};
解题思路:这道题还是利用dfs算法解决,通过递归遍历得到最终的结果。
代码实现:
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int sumNumbers(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int sum = 0, temp = 0;
if(root == NULL)
{
return 0;
}
search(root, sum, temp);
return sum;
}
void search(TreeNode *p, int &sum, int temp)
{
if(p->left == NULL && p->right == NULL)
{
temp = temp * 10 + p->val;
sum += temp;
return;
}
temp = temp * 10 + p->val;
if(p->left != NULL)
{
search(p->left, sum, temp);
}
if(p->right != NULL)
{
search(p->right, sum, temp);
}
}
};
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