Sum Root to Leaf Numbers

最新推荐文章于 2024-08-27 16:38:35 发布

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本文探讨了如何计算二叉树中所有从根节点到叶子节点路径所代表的数字之和。通过两种方法实现：一种使用堆栈进行后续遍历，另一种采用递归帮助函数。示例展示了输入为特定二叉树时的计算过程和结果。

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题目：

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.

解法1：

此题可以用二叉树的后续遍历

/**
 * 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 sumStack(stack<TreeNode*> s)
    {
        int sum = 0;
        int i = 0;
        while(!s.empty())
        {
            sum = sum + s.top()->val * (int)pow(10,i);
            ++i;
            s.pop();
        }
        return sum;
    }
    int sumNumbers(TreeNode* root) {
        int sum = 0;
        stack<TreeNode* > s;
        TreeNode* p = nullptr;
        while(root != nullptr || !s.empty())
        {
            if(root != nullptr)
            {
                s.push(root);
                root = root->left;
            }
            else
            {
                root = s.top();
                if(root->left == nullptr && root->right == nullptr)
                {
                    sum += sumStack(s);
                }

                if(root->right != nullptr && root->right != p)
                {
                    root = root->right;
                }
                else
                {
                    s.pop();
                    p = root;
                    root = nullptr;                 
                }                   
            }
        }
        return sum;
    }
};

解法2：

class Solution {
public:
    int sumNumbersHelp(TreeNode* root, int val)
    {
        if(root == nullptr)
            return 0;
        val = 10 * val + root->val;
        if(root->left == nullptr && root->right == nullptr)
            return val;
        return sumNumbersHelp(root->left, val) + sumNumbersHelp(root->right, val);
    }
    int sumNumbers(TreeNode* root) {
        return sumNumbersHelp(root, 0);
    }
};