129. Sum Root to Leaf Numbers

最新推荐文章于 2020-02-04 14:57:49 发布

万里归来少年心

最新推荐文章于 2020-02-04 14:57:49 发布

阅读量227

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode 文章标签：遍历二叉树路径和递归搜索

本文链接：https://blog.youkuaiyun.com/liyazhen2011/article/details/83274181

LeetCode 专栏收录该内容

85 篇文章

订阅专栏

本文探讨了如何求解二叉树中所有根到叶子节点路径的数字总和，通过递归与非递归两种算法实现，展示了清晰的代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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->2represents 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.

思路1：递归

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

思路2：非递归

使用栈对遍历过得节点进行存储，以方便节点寻找上一级节点。

/**
 * 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) {
       stack<TreeNode*> nodes;
        if (!root) return 0;
        
        int total = 0;
        int current = 0;
        TreeNode *last = nullptr;
        while (root || !nodes.empty()) {
            if (root) {
                nodes.push(root);
                current = current*10 + root->val;
                root = root->left;
            } else {
                root = nodes.top();
                if (root->right && root->right != last) {
                    root = root->right;
                } else {
                    nodes.pop();
                    last = root;
                    if (root->left == nullptr && root->right == nullptr)
                        total += current;
                    current /= 10;
                    root = nullptr;
                }
            }
        }
        return total;
    }

};