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;
}
};
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