Sum Root to Leaf Numbers @LeetCode
PROBLEM :
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.
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SIMPLE c++ IMPLEMENTATION :
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/**
* 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) {
if(root==NULL)
return 0 ;
if(root->left==NULL&&root->right==NULL)
return root->val ;
return SumNumPath(root->left,root->val)+ SumNumPath(root->right,root->val) ;
}
int SumNumPath(TreeNode *root,int ele){
if(root==NULL)
return 0 ;
if(root->left==NULL&&root->right==NULL)
return ele*10+root->val ;
int l,r ;
l=0,r=0 ;
if(root->left)
l= SumNumPath(root->left,ele*10+root->val) ;
if(root->right)
r= SumNumPath(root->right,ele*10+root->val) ;
return l+r ;
}
};
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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.
--------------------------------------------------------------------------------
SIMPLE c++ IMPLEMENTATION :
--------------------------------------------------------------------------------
/**
* 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) {
if(root==NULL)
return 0 ;
if(root->left==NULL&&root->right==NULL)
return root->val ;
return SumNumPath(root->left,root->val)+ SumNumPath(root->right,root->val) ;
}
int SumNumPath(TreeNode *root,int ele){
if(root==NULL)
return 0 ;
if(root->left==NULL&&root->right==NULL)
return ele*10+root->val ;
int l,r ;
l=0,r=0 ;
if(root->left)
l= SumNumPath(root->left,ele*10+root->val) ;
if(root->right)
r= SumNumPath(root->right,ele*10+root->val) ;
return l+r ;
}
};
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