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                合規國際互聯網加速 OSASE為企業客戶提供高速穩定SD-WAN國際加速解決方案。 廣告
                **一. 題目描述** 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`. **二. 題目分析** 這道題只是一道二叉樹的深度優先搜索的題目,在葉結點時將從根到葉結點的路徑上的結點的值組成一個十進制的數,本質上還是一道深度優先搜索的題。 **三. 示例代碼** ~~~ #include <iostream> struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; class Solution { void sumNumbers(TreeNode*root, int &Result, int &TmpResult) { if (!root) { return; } // update the TmpResult TmpResult = TmpResult * 10 + root->val; if (!root->left && !root->right) { Result += TmpResult; } if (root->left) { sumNumbers(root->left, Result, TmpResult); } if (root->right) { sumNumbers(root->right, Result, TmpResult); } // restore the TmpResult; TmpResult = (TmpResult - root->val) / 10; } public: int sumNumbers(TreeNode* root) { int Result = 0; int TmpResult = 0; sumNumbers(root, Result, TmpResult); return Result; } }; ~~~ **四. 小結** 這道題是一道二叉樹深度優先搜索的變形題目,考察的還是二叉樹的遞歸遍歷,只是換了一種考察方式而已。
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