Sum root to leaf numbers -- leetcode

Given a Binary Tree Containing digits from0-9Only, each root-to-leaf path cocould represent a number.

An example is the root-to-leaf path1->2->3Which represents the number123.

Find the total sum of all root-to-leaf numbers.

For example,

    1   /   2   3

The root-to-leaf path1->2Represents the number12.
The root-to-leaf path1->3Represents the number13.

Return the sum = 12 + 13 =25.

Basic Ideas:

Pre-order traversal.

Use a global variable to record and.

Each time a leaf node receives a number, it accumulates and coexist with global variables.

The actual running time on leetcode is 4 ms.

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int sumNumbers(TreeNode* root) {        int sum = 0;        sumNumbers(root, 0, sum);        return sum;    }        void sumNumbers(TreeNode* root, int number, int &sum) {        if (!root) return;        number = number * 10 + root->val;        if (!root->left && !root->right)            sum += number;        sumNumbers(root->left, number, sum);        sumNumbers(root->right, number, sum);    }};

Ultimate Edition:

You can use the function return value to replace the above global variables and (the third variable of the function ).

If it is a page subnode, an integer is returned.

If it is an intermediate node, return the integer and between its left and right subnumbers.

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

Sum root to leaf numbers -- leetcode