leetcode 129. Sum Root to Leaf Numbers-优快云博客

本文链接：https://blog.youkuaiyun.com/level_code/article/details/97611581

本文介绍了一种算法，用于解决给定二叉树中，所有从根节点到叶子节点路径所代表数字的总和问题。通过递归方式，每深入一层就将当前数字乘以10加上节点值，直至到达叶子节点返回最终数值。

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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]

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.

给出一个二叉树，每个root到leaf的路径代表一个数字，问所有路径的数字和是多少

思路：
每往下一level走一步，就需要前一数字10 + 正在访问点的val
假设上一层的值是prefix，那么正在访问节点处的值就是prefix10 + root.val
而且左右子树共用同一个prefix

当已经是leaf时，返回prefix* 10 + root.val即可

//0ms
    public int sumNumbers(TreeNode root) {
        return helper(root, 0);
    }
    
    public int helper(TreeNode root, int prefix) {
        if (root == null) {
            return 0;
        }
        
        int sum = prefix * 10 + root.val;
        
        if (root.left == null && root.right == null) {
            return sum;
        }
        
        return helper(root.left, sum) + helper(root.right, sum);
    }