[LeetCode] Sum Root to Leaf Numbers

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.

思路：

典型的root->leaf path问题，遍历所有path并更新sum。传递的变量是当前节点的path num值curNum。则下一层的节点对应的path num为curNum*10 + val。

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

class Solution { public: int sumNumbers(TreeNode *root) { int sum = 0; sumRootToLeafNum(root, 0, sum); return sum; } void sumRootToLeafNum(TreeNode *root, int curNum, int &sum) { if(!root) return; curNum = curNum*10 + root->val; if(!root->left && !root->right) sum += curNum; if(root->left) sumRootToLeafNum(root->left, curNum, sum); if(root->right) sumRootToLeafNum(root->right, curNum, sum); } };

总结：

当root->leaf path很长时，path num将会很大，sum的时候很容易出现溢出。需要和面试官讨论是否要特殊处理这样的情况。另外当root为NULL时需要返回0.