129. Sum Root to Leaf Numbers-优快云博客

本文探讨了如何在仅包含0-9数字的二叉树中寻找所有从根节点到叶子节点路径所代表的数字总和。通过遍历二叉树，采用递归算法累积路径数值，最终得出所有可能路径的总和。

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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.

/**
 * 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;
        Sum=0;
        deepSum(0,root);
        return Sum;
    }
    void deepSum(int CurrentSum, TreeNode* &root){
        CurrentSum=CurrentSum*10+root->val;
        if(root->left==NULL&&root->right==NULL){
            Sum+=CurrentSum;
        }
        if(root->left)
            deepSum(CurrentSum,root->left);
        if(root->right)
            deepSum(CurrentSum,root->right);
    }
private:
    int Sum;
};