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   3Output: 25Explanation: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   1Output: 1026Explanation: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;    }};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/129

类似题目:

Path Sum

Binary Tree Maximum Path Sum

参考资料:

https://leetcode.com/problems/sum-root-to-leaf-numbers/

https://leetcode.com/problems/sum-root-to-leaf-numbers/discuss/41367/Non-recursive-preorder-traverse-Java-solution

https://leetcode.com/problems/sum-root-to-leaf-numbers/discuss/41452/Iterative-C%2B%2B-solution-using-stack-(similar-to-postorder-traversal)

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C++实现LeetCode(129.求根到叶节点数字之和)

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