[leetcode] Sum Root to Leaf Numbers

最新推荐文章于 2022-01-25 11:20:00 发布

Cabinathor

最新推荐文章于 2022-01-25 11:20:00 发布

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分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/lmxmimihuhu/article/details/46641287

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本文介绍了一种算法，用于计算二叉树中所有从根节点到叶子节点的路径所代表数字的总和。通过递归遍历的方式，将每个路径上的数字累加并返回最终的总和。

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From : https://leetcode.com/problems/sum-root-to-leaf-numbers/

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.

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

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private int sum = 0;
    public int sumNumbers(TreeNode root) {
        dfs(root, 0);
        return sum;
    }
    
    void dfs(TreeNode root, int a) {// 当前root待计算，已累计到a
        if(root == null) {
            return;
        }
        a = a*10+root.val;
        if(root.left == null && root.right == null) {
            sum += a;
            return;
        }
        dfs(root.left, a);
        dfs(root.right, a);
    }
}