C++实现LeetCode(129.求根到叶节点数字之和)

[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;

									    }

									};