84:Sum Root to Leaf Numbers
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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.
解析1:遍历出树的所有路径,然后将这些路径的和相加,该解析方法和 path sum ii 那道题目解析2方法很像(http://blog.csdn.net/weishenmetlc/article/details/61209222),因为都要遍历出所有路径, 代码如下:
// 递归法,时间复杂度 O(n),空间复杂度 O(logn)class Solution {public: int sumNumbers(TreeNode* root) { int sum = 0; int tmp = 0; sumNumbers(root, tmp, sum); return sum; }private: void sumNumbers(TreeNode* root, int& tmp, int& sum) { if (!root) return; tmp = tmp * 10 + root -> val; if (!root -> left && !root -> right) sum += tmp; sumNumbers(root -> left, tmp, sum); sumNumbers(root -> right, tmp, sum); tmp /= 10; }};
解析2:先求出从根节点到其左子树的所有路径和,再求出从根节点到其右子树的所有路径和,相比如解析1, 该方法更易理解,代码如下:
下面代码的思想及编写参考了网址https://github.com/soulmachine/leetcode#leetcode题解题目
// 递归法,时间复杂度 O(n),空间复杂度 O(logn)class Solution {public: int sumNumbers(TreeNode* root) { return dfs(root, 0); }private: int dfs(TreeNode* root, int sum) { if (!root) return 0; if (!root -> left && !root -> right) return 10 * sum + root -> val; return dfs(root -> left, sum * 10 + root -> val) + dfs(root -> right, sum * 10 + root -> val); }};
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