129. Sum Root to Leaf Numbers

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129. Sum Root to Leaf Numbers

题目

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. 

解析

• 先序遍历的思想(根左右)+数字求和(每一层都比上层和*10+当前根节点的值)
• 可以从上到下累加，直到叶子节点，然后累加！

class Solution_129 {
public:
	int dfs(TreeNode* root, int sum)
	{
		if (!root)
		{
			return;
		}
		sum = sum * 10 + root->val;
		if (root->left==NULL&&root->right==NULL)
		{
			return sum;
		}

		return dfs(root->left, sum) + dfs(root->right, sum);
	}

	int sumNumbers_ref(TreeNode* root)
	{
		if (root == NULL)
		{
			return 0;
		}
		int sum = 0;

		return dfs(root, sum); //从根节点开始
	}

	int sumNumbers(TreeNode *root) {
		if (!root)
		{
			return 0;
		}
		stack<TreeNode*> sta;
		sta.push(root);

		int ret = 0;
		TreeNode* top = 0;
		while (!sta.empty())
		{
			top = sta.top();
			sta.pop();

			if (!top->left && !top->right)
			{
				ret += top->val;
			}
			if (top->left)
			{
				top->left->val += 10 * top->val; //但是这样改变了节点的值，可以将TreeNode* 和累积和(一个变量)组成一个pair，分开处理
				sta.push(top->left);
			}
			if (top->right)
			{
				top->right->val += 10 * top->val;
				sta.push(top->right);
			}
		}
		return ret;
	}
};

题目来源

• 129. Sum Root to Leaf Numbers
• 尾递归=递归+迭代？