129. Sum Root to Leaf Numbers (A good example to compare to Styles to solve tree problems)_shortest root to leaf path sum equals to a given n-优快云博客

本文介绍了一种算法，用于求解二叉树中所有从根节点到叶子节点的路径所表示数字的总和。通过两种不同的实现方式——使用全局变量的深度优先搜索和分治法，展示了如何有效地解决这个问题。

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.

The first one using a global variable and do dfs.

Code:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    int ret = 0;
    public int sumNumbers(TreeNode root) {
        if(root == null) return 0;
        helper(0,root);
        return ret;
    }
    
    void helper(int path, TreeNode root){
        path *= 10;
        path += root.val;
        if(root.left == null && root.right == null){
            ret += path;
            return;
        }
        if(root.left!=null){
            helper(path,root.left);
        }
        if(root.right!=null){
            helper(path,root.right);
        }
    }
}

The second one, divide and conquer

public class Solution {
    public int sumNumbers(TreeNode root) {
        return helper(0,root);
    }
    
    int helper(int path, TreeNode root){
        if(root == null) return 0;
        path *= 10;
        path += root.val;
        if(root.left == null && root.right == null){
            return path;
        }
        int ans = 0;
        if(root.left!=null){
            ans += helper(path,root.left);
        }
        if(root.right!=null){
            ans += helper(path,root.right);
        }
        return ans;
    }
}