129. Sum Root to Leaf Numbers

2024年1月27日 199次阅读

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.

方法1:遍历，依次改变每一个节点的值(前一个节点*10加上当前节点值)，当为叶子节点时将当前值添加到总值中。

/**
 * 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;
        int sum=0;
        ds(root,sum,0);
        return sum;    
    }
    void ds(TreeNode* root,int &sum,int preVal){
        root->val=preVal*10+root->val;
        if(root->left==NULL&&root->right==NULL) sum+=root->val;
        if(root->left!=NULL) ds(root->left,sum,root->val);
        if(root->right!=NULL) ds(root->right,sum,root->val);
    }
};

 

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