[LeetCode] 113: Sum Root to Leaf Numbers

最新推荐文章于 2023-10-10 09:42:58 发布

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本文介绍了一种解决特定二叉树问题的方法——求所有从根节点到叶子节点的路径数值之和。通过递归算法，该解决方案遍历每个可能的路径并计算其总和。

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[Problem]

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.

[Solution]

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
	vector<int> generatePaths(TreeNode *root, int preVal){
		vector<int> res;
		if(root == NULL){
			return res;
		}
		else if(root->left == NULL && root->right == NULL){
			res.push_back(preVal*10 + root->val);
		}
		else{
			vector<int> left = generatePaths(root->left, preVal*10 + root->val);
			vector<int> right = generatePaths(root->right, preVal*10 + root->val);
			for(int i = 0; i < left.size(); ++i){
				res.push_back(left[i]);
			}
			for(int i = 0; i < right.size(); ++i){
				res.push_back(right[i]);
			}
		}
		
		return res;
	}
	int sumNumbers(TreeNode *root) {
		// Note: The Solution object is instantiated only once and is reused by each test case.
		vector<int> paths = generatePaths(root, 0);
		int res = 0;
		for(int i = 0; i < paths.size(); ++i){
			res += paths[i];
		}
		return res;
	}
};