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 = ;
Helper(root, , 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 += (*partSum+root->val);
else
{
Helper(root->left, *partSum+root->val, Sum);
Helper(root->right, *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 ;
int ret = ;
int cur = ;
stack<TreeNode*> stk;
unordered_map<TreeNode*, bool> 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* + top->left->val;
continue;
}
if(top->right != NULL && visited[top->right] == false)
{
stk.push(top->right);
visited[top->right] = true;
cur = cur* + top->right->val;
continue;
}
if(top->left == NULL && top->right == NULL)
{
ret += cur;
}
stk.pop();
cur = (cur - top->val) / ;
}
return ret;
}
};

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