Leetcode--Sum Root to Leaf Numbers
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Problem Description:
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
.
分析:
题目本身看起来很简单的,但是如果对于对属性结构和递归编程不是那么熟的人来说就会遇到很细节的问题了。比如:
1. 递归到该点的时候该如何操作?
2. 由跟节点到叶子节点的时候,每进入一层就需要把前面的值*10,如何处理这个值?总值又是如何保存的呢?
问题处理:
1、 分情况就是:一、 空节点, 二、叶子节点, 三、 一般节点(包括只有左孩子,只有右孩子,两个孩子都有),递归到该点的时候就把这点的值与前面的值乘以10,然后相加。
2、 需要如何保存前面的值?以参数传递的方式保存,以供下一层使用。可以以两种方式保存:参数传递方式,返回函数值方式。
代码如下:
/** * 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 leafnum(TreeNode *root,int num) { if(!root) return 0; if(!root->left&&!root->right) return num*10+root->val; else return leafnum(root->left,num*10+root->val)+leafnum(root->right,num*10+root->val); } int sumNumbers(TreeNode *root) { int sum=0; if(!root) return 0; if(!root->left&&!root->right) return root->val; sum=leafnum(root->left,root->val)+leafnum(root->right,root->val); return sum; }};
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