Sum Root to Leaf Numbers

最新推荐文章于 2020-02-27 00:23:14 发布

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本文探讨了一个关于二叉树的算法问题：如何计算所有从根节点到叶子节点的路径数值之和。通过递归的方法解决该问题，并分享了解题过程中遇到的陷阱及解决方案。

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.

这个题挺简单，但是我还是出了一点问题，一开始我以为第归终止简单的就是root == NULL，提交了发现这样会重复计数，因为每个叶结点有两个空子节点。大意了。

class Solution {
private:
	int sum(TreeNode *root, int pre)
	{
		if(root == NULL)
		{
			return pre;
		}
		int that = pre * 10 + root->val;
		if(root->left == NULL)
		{
			return sum(root->right, that);
		}
		else if(root->right == NULL)
		{
			return sum(root->left, that);
		}
		else
		{
			return sum(root->left, that) + sum(root->right, that);
		}
	}
public:
    int sumNumbers(TreeNode *root) {
        return sum(root, 0);
    }
};