【Sum Root to Leaf Numbers】cpp
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
.
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) { return Solution::sumFromRoot2Leaf(root, 0); } static int sumFromRoot2Leaf(TreeNode* root, int upperVal) { if ( !root ) return 0; if ( !root->left && !root->right ) return upperVal*10+root->val; return Solution::sumFromRoot2Leaf(root->left, upperVal*10+root->val) + Solution::sumFromRoot2Leaf(root->right, upperVal*10+root->val); } };
tips:
一开始想自低向上算,后来发现自顶向下算,每次传入上一层的值比较科学。
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第二次过这道题,一次AC了,沿用dfs写法。
/** * 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 ret = 0; if ( root ) Solution::dfs(root, 0, ret); return ret; } static void dfs(TreeNode* root, int sum, int& ret ) { if ( !root->left && !root->right ) { sum = sum*10 + root->val; ret += sum; return; } if ( root->left ) Solution::dfs(root->left, sum*10+root->val, ret); if ( root->right ) Solution::dfs(root->right, sum*10+root->val, ret); } };