leetcode question 129: Sum Root to Leaf Numbers-优快云博客

本文探讨了如何在二叉树中寻找所有根节点到叶子节点的数字路径，并计算这些路径所代表数字的总和。通过递归算法，文章详细解释了如何遍历二叉树的每一个可能路径，将路径上的数字连接起来形成完整的数字，并最终求得所有路径数字的总和。

问题：

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.

分析：

这道题叫我们找出所有的由根节点到子节点的数字组成的数，并求和。我们就直接分两步走，先是找出所有的数，存到数组里，然后再把所有的数求和。比较注意的就是，需要注意考虑到达当前节点时，当前节点为空、只有一个子树等等的情况，综合判断。

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) {
        if( root == NULL ){
            return 0;
        }
        //找到所有的数
        list<int> ans = findNum(root,0);
        int sum = 0;
        for( list<int>::iterator i = ans.begin() ; i != ans.end() ; i ++ ){
            sum += *i;
        }
        return sum;
    }
    
    list<int> findNum( TreeNode* treePointer , int val ){
        list<int> num;
        
        //如果是空则返回空的list
        if ( treePointer == NULL ){
            return num;
        }
        
        //是子节点就返回
        if( treePointer->left == NULL && treePointer->right == NULL ){
            num.push_back( val*10 + treePointer->val );
            return num;
        }
        
        //向左子树和右子树分别往下找
        list<int> leftNum = findNum(treePointer->left,val*10 + treePointer->val);
        list<int> rightNum = findNum(treePointer->right,val*10 + treePointer->val);
        
        //合并两个子树返回的list
        if(leftNum.size() == 0 ){
            return rightNum;
        }
        if(rightNum.size() == 0 ){
            return leftNum;
        }
        
        leftNum.splice(leftNum.begin(),rightNum);
        return leftNum;
    }
};