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.

 

每到达一个根节点，就代表一个路径访问完成，将和加入总和。

 

解法一：

树结构用递归是最容易的，每递归一层，上层的部分和需要乘以10在做加法。

/** * Definition for binary tree * 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;        Helper(root, 0, Sum);        return Sum;    }    void Helper(TreeNode* root, int partSum, int& Sum)    {        if(root == NULL)            return;        else if(root->left == NULL && root->right == NULL)  //add this path            Sum += (10*partSum+root->val);        else        {            Helper(root->left, 10*partSum+root->val, Sum);            Helper(root->right, 10*partSum+root->val, Sum);        }    }};

 

解法二：

非递归方法，使用栈存放当前的路径进行深度搜索，并设置部分和记录栈中的数值。

结合进栈与出栈进行部分和调整：

进栈：部分和*10+当前值

出栈：(部分和-当前值)/10

如果遍历到叶节点，则将部分和加入总和。

/** * 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) {        if(root == NULL)            return 0;        int ret = 0;        int cur = 0;        stack
     
       stk;        unordered_map
      
        visited;        stk.push(root);        visited[root] = true;        cur += root->val;        while(!stk.empty())        {            TreeNode* top = stk.top();            if(top->left != NULL && visited[top->left] == false)            {                stk.push(top->left);                visited[top->left] = true;                cur = cur*10 + top->left->val;                continue;            }            if(top->right != NULL && visited[top->right] == false)            {                stk.push(top->right);                visited[top->right] = true;                cur = cur*10 + top->right->val;                continue;            }            if(top->left == NULL && top->right == NULL)            {                ret += cur;            }            stk.pop();            cur = (cur - top->val) / 10;        }        return ret;    }};