[leetcode]129. Sum Root to Leaf Numbers@Java解题报告

最新推荐文章于 2024-06-04 05:35:32 发布

JacobGo

最新推荐文章于 2024-06-04 05:35:32 发布

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本文详细解析了LeetCode上的第129题——求根节点到叶节点的所有路径之和。针对该二叉树问题，提供了两种不同的解决方案，并通过具体的代码实现展示了如何递归地计算所有根节点到叶节点路径所代表的数字之和。

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

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.

package go.jacob.day810;

/**
 * 129. Sum Root to Leaf Numbers
 * 
 * @author Jacob
 *
 */
public class Demo1 {
	public int sumNumbers(TreeNode root) {
		if (root == null)
			return 0;
		int sum = 0;
		return sumNumbers(root, sum);
	}

	private int sumNumbers(TreeNode root, int sum) {
		if (root == null)
			return 0;
		if (root.left == null && root.right == null)
			return sum * 10 + root.val;
		return sumNumbers(root.left, 10 * sum + root.val) + 
				sumNumbers(root.right, 10 * sum + root.val);

	}
	/*
	 * My Solution
	 */
	private int sumNumbers_1(TreeNode root, int sum) {
		sum += root.val;
		int leftSum = 0;
		int rightSum = 0;
		if (root.left != null)
			leftSum = sumNumbers(root.left, 10 * (sum));
		if (root.right != null)
			rightSum = sumNumbers(root.right, 10 * (sum));
		// 如果root的左右子树为空，返回sum。否则返回左右子树递归的结果和
		return root.left == null && root.right == null ? sum : leftSum + rightSum;

	}
}