Task
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.
Note: A leaf is a node with no children.
Example
Example 1:
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Input: [1,2,3]
1
/ \
2 3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.
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Example 2:
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Input: [4,9,0,5,1]
4
/ \
9 0
/ \
5 1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.
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Solution
- 维护两个计数：
- num记录每条路径的数值，它不仅仅是一个变量，在每次分叉时都需要复制一份。使用递归实现时，即是每次分叉时将num传入递归的函数（拷贝）
- ans记录最终的结果数值，当遍历到叶子节点（无左右子节点）时就将这条路径的num加到ans上
- 时间复杂度：每个节点看一次，O(n)
- 空间复杂度：取决于树的高度，O(h)
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// Runtime: 4 ms, faster than 82.80% of C++ online submissions for Sum Root to Leaf Numbers.
// Memory Usage: 12.6 MB, less than 38.01% of C++ online submissions for Sum Root to Leaf Numbers.
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int sumNumbers(TreeNode* root) {
traverse(root,0);
return ans;
}
private:
void traverse(TreeNode *root,int num){
//递归终止
if(!root) return;
//更新num，用于记录每一条路径的数值
num=num*10+root->val;
//当有左子树或右子树时，将num复制一份分别传入左右子树
if(root->left || root->right){
traverse(root->left,num);
traverse(root->right,num);
}
//遍历到叶子节点，更新ans
else{
ans+=num;
}
}
//成员变量，类内初始化为0
int ans=0;
};
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