劍指 Offer II 049. 從根節點到葉節點的路徑數字之和
阿新 • • 發佈:2021-10-07
dfs水題
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * };自己選擇的路,跪著也要走完。朋友們,雖然這個世界日益浮躁起來,只要能夠為了當時純粹的夢想和感動堅持努力下去,不管其它人怎麼樣,我們也能夠保持自己的本色走下去。*/ class Solution { public: int sum = 0; void dfs(TreeNode* root, int num) { num = num * 10 + root->val; if(root->left == nullptr && root->right == nullptr) { sum += num; return; } if(root->left) { dfs(root->left, num); } if(root->right) dfs(root->right, num); num = (num - root->val) / 10; } int sumNumbers(TreeNode* root) { dfs(root, sum); return sum; } };