129. Sum Root to Leaf Numbers

最新推荐文章于 2020-12-22 20:22:53 发布

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最新推荐文章于 2020-12-22 20:22:53 发布

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本文介绍了一种算法问题，即给定一个只包含0到9数字的二叉树，每条从根节点到叶子节点的路径可以看作一个数字。文章详细解释了如何寻找所有这样的路径并计算它们所代表的数字之和。

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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.

题意：路径组成数字，计算这些路径的和。

思路：找出各个路径的数字，然后求和。

/**
 * 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;
		getPathString(root, "", sum);
		return sum;
	}
private:
	void getPathString(TreeNode* root, string cur, int& sum){
		if (root->left == NULL  && root->right == NULL){
			cur += to_string(root->val);
			sum += atoi(cur.c_str());
			return;
		}
		if (root->left){
			getPathString(root->left, cur + to_string(root->val), sum);
		}
		if (root->right){
			getPathString(root->right, cur + to_string(root->val), sum);
		}
		return;
	}
};

思路2：上述思路路径信息的保存用的是string，以下思路路径信息用int值存储。

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