LeetCode六月挑战（6.26 ）Sum Root to Leaf Numbers LeetCode 129解题方案

昨天的题只需要双重循环即可完成，很简单所以没有专门发一个

题目描述

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.

思路

如图， 递归节点，如果此节点是根节点，判断是否为null，如果为null，则返回0，接着判断他的左右子树是不是空，如果为空直接返回当前值，如果不为空，递归左右子树。

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int sumNumbers(TreeNode root) {
        return findNode(root,0);
    }
    
    public int findNode(TreeNode root, int n){
        if(root == null) return 0;
        int cur = root.val + n * 10;
        if(root.left == null && root.right == null) return cur;
        return findNode(root.left,cur)+ findNode(root.right,cur);
    }
}