[LeetCode] Sum Root to Leaf Numbers

最新推荐文章于 2022-01-25 11:20:00 发布

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最新推荐文章于 2022-01-25 11:20:00 发布

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本文介绍了一种算法，用于计算二叉树中从根节点到叶子节点的所有路径所代表的数字之和。通过深度优先搜索策略遍历二叉树，并累积计算路径数值。

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.

求所有根到叶子节点组成数字之和

思路很直观，深搜到所有叶结点，加到结果sum中

/**
 * 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 sumNumbers(TreeNode *root) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
       if(root == NULL) return 0;
       int totalSum = 0;
       int currentSum = 0;
       dfs(root,currentSum,totalSum);
       return totalSum;
    }
    
    void dfs(TreeNode *root,int currentSum, int &totalSum) {
        currentSum = currentSum*10 + root->val;
        if(root->left == NULL && root->right == NULL) {
            // this is a leaf node add sum to total
            totalSum += currentSum;
            return ;
        }
        if(root->left != NULL) {
            dfs(root->left,currentSum,totalSum);
        }
        if(root->right != NULL) {
            dfs(root->right,currentSum,totalSum);
        }
    }
};