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
.
从根节点開始,dfs的思路,事实上也就是postorder(先序遍历),遍历路径上的值每次乘以基数10,过程非常easy,代码例如以下:
class Solution { public: int sum; void dfs(TreeNode *root, int pre){ if (root == NULL) return; int current = root->val + 10 * pre; if (root->left == NULL && root->right == NULL){ sum += current; return; } if (root->left) dfs(root->left, current); if (root->right) dfs(root->right, current); } int sumNumbers(TreeNode *root) { sum = 0; dfs(root, 0); return sum; } };