[leetcode] 129. Sum Root to Leaf Numbers-优快云博客

本文链接：https://blog.youkuaiyun.com/w5688414/article/details/97365368

本文探讨了一道经典的二叉树遍历问题，即求所有从根节点到叶子节点路径所组成的数字之和。通过先序遍历算法，递归地计算每条路径的数值并累加得到最终结果。提供了C++和Python两种语言的实现代码。

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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.

Note: A leaf is a node with no children.

Example:

Input:

[1,2,3]
    1
   / \
  2   3

Output:

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:

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.

分析

题目的意思是：给定一棵二叉树，从根节点到叶子节点组成一个数字，然后求和。

先序遍历每个结点，用一个sum变量保存求和的值，每一层都比上层和*10+当前根节点的值。

代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sumNumbers(TreeNode* root) {
        int sum=0;
        int res=0;
        solve(root,sum,res);
        return res;
    }
    void solve(TreeNode* root, int sum,int& res){
        if(!root){
            return ;
        }
        sum=sum*10+root->val;
        if(!root->left&&!root->right){
            res+=sum;
            return ;
        }
        solve(root->left,sum,res);
        solve(root->right,sum,res);
    }
};

代码二

牛课网也有类似的题目，我用python重新实现了一下，发现好久不练都不会了哈哈。

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

#
# 
# @param root TreeNode类 
# @return int整型
#
class Solution:
    def dfs(self,root,val):
        if(root is None):
            return
        cur=val*10+root.val
        if(root.left is None and root.right is None):
            self.nums.append(cur)
        self.dfs(root.left,cur)
        self.dfs(root.right,cur)
        
    def sumNumbers(self , root ):
        # write code here
        self.nums=[]
        self.dfs(root,0)
        return sum(self.nums)