129. Sum Root to Leaf Numbers (M)

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本文深入探讨了一种经典的回溯法算法，用于解决从树的根节点到叶节点路径上数字组合成整数并求和的问题。通过两个递归方法的详细解释，帮助读者理解如何高效地遍历树结构，计算所有可能路径的总和。

Sum Root to Leaf Numbers (M)

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.

Note: A leaf is a node with no children.

Example:

Input: [1,2,3]
    1
   / \
  2   3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.

Example 2:

Input: [4,9,0,5,1]
    4
   / \
  9   0
 / \
5   1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.

题意

将树从根节点到叶结点路径上的所有数字组合成一个整数，求所有这样的整数的和。

思路

典型的回溯法。

代码实现 - 递归 1

class Solution {
    int sum = 0;

    public int sumNumbers(TreeNode root) {
        if (root != null) {
            dfs(root, 0);
        }

        return sum;
    }

    private void dfs(TreeNode x, int cur) {
        if (x.left == null && x.right == null) {
            sum += cur * 10 + x.val;
            return;
        }

        if (x.left != null) {
            dfs(x.left, cur * 10 + x.val);
        }
        if (x.right != null) {
            dfs(x.right, cur * 10 + x.val);
        }
    }
}

代码实现 - 递归 2

class Solution {
    public int sumNumbers(TreeNode root) {
        return dfs(root, 0);
    }

    private int dfs(TreeNode x, int sum) {
        if (x == null) {
            return 0;
        }

        sum = sum * 10 + x.val;
        
        if (x.left == null && x.right == null) {
            return sum;
        }

        return dfs(x.left, sum) + dfs(x.right, sum);
    }
}