LeetCode :: Sum Root to Leaf Numbers [tree、dfs]

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

.

从根节点开始，dfs的思路，其实也就是postorder（先序遍历），遍历路径上的值每次乘以基数10，过程很简单，代码如下：

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;
    }
};