Sum Root to Leaf Numbers
Sum Root to Leaf Numbers
题目
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.
思路
DFS dfs(root,root->val)
解题
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24/**
* 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 sum;
int sumNumbers(TreeNode* root) {
if(root==NULL) return 0;
sum=0;
dfs(root,root->val);
return sum;
}
void dfs(TreeNode *root,int num){
if(root->left==NULL && root->right==NULL) sum+=num;
if(root->left) dfs(root->left,num*10+root->left->val);
if(root->right) dfs(root->right,num*10+root->right->val);
}
};
python 版1
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26# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# @param {TreeNode} root
# @return {integer}
def __init__(self):
self.sum=0
def sumNumbers(self, root):
if None == root:
return 0
self.dfs(root, root.val)
return self.sum
def dfs(self, root, val):
if not root.left and not root.right:
self.sum += val
if None != root.left:
self.dfs(root.left, val * 10 + root.left.val)
if None != root.right:
self.dfs(root.right, val * 10 + root.right.val)
