C++中怎么利用LeetCode求根到叶节点数字之和
C++中怎么利用LeetCode求根到叶节点数字之和,很多新手对此不是很清楚,为了帮助大家解决这个难题,下面小编将为大家详细讲解,有这方面需求的人可以来学习下,希望你能有所收获。
[LeetCode] 129. Sum Root to Leaf Numbers 求根到叶节点数字之和
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:
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
.
Example 2:
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
.
这道求根到叶节点数字之和的题跟之前的求 Path Sum 很类似,都是利用DFS递归来解,这道题由于不是单纯的把各个节点的数字相加,而是每遇到一个新的子结点的数字,要把父结点的数字扩大10倍之后再相加。如果遍历到叶结点了,就将当前的累加结果sum返回。如果不是,则对其左右子结点分别调用递归函数,将两个结果相加返回即可,参见代码如下:
解法一:
class Solution { public: int sumNumbers(TreeNode* root) { return sumNumbersDFS(root, 0); } int sumNumbersDFS(TreeNode* root, int sum) { if (!root) return 0; sum = sum * 10 + root->val; if (!root->left && !root->right) return sum; return sumNumbersDFS(root->left, sum) + sumNumbersDFS(root->right, sum); } };
我们也可以采用迭代的写法,这里用的是先序遍历的迭代写法,使用栈来辅助遍历,首先将根结点压入栈,然后进行while循环,取出栈顶元素,如果是叶结点,那么将其值加入结果res。如果其右子结点存在,那么其结点值加上当前结点值的10倍,再将右子结点压入栈。同理,若左子结点存在,那么其结点值加上当前结点值的10倍,再将左子结点压入栈,是不是跟之前的 Path Sum 极其类似呢,参见代码如下:
解法二:
class Solution { public: int sumNumbers(TreeNode* root) { if (!root) return 0; int res = 0; stack<TreeNode*> st{{root}}; while (!st.empty()) { TreeNode *t = st.top(); st.pop(); if (!t->left && !t->right) { res += t->val; } if (t->right) { t->right->val += t->val * 10; st.push(t->right); } if (t->left) { t->left->val += t->val * 10; st.push(t->left); } } return res; } };
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