GuilinDev

Lc0129

05 August 2008

129 Sum Root to Leaf Numbers

原题概述

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:

1
2
3
4
5
6
7
8
9
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:

1
2
3
4
5
6
7
8
9
10
11
12
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.

题意和分析

这是一道树的题目,一般使用递归来做,主要就是考虑递归条件和结束条件。这道题思路还是比较明确的,目标是把从根到叶子节点的所有路径得到的整数都累加起来,递归条件即是把当前的sum乘以10并且加上当前节点传入下一函数,进行递归,最终把左右子树的总和相加。结束条件的话就是如果一个节点是叶子,那么我们应该累加到结果总和中,如果节点到了空节点,则不是叶子节点,不需要加入到结果中,直接返回0即可。算法的本质是一次先序遍历,所以时间是O(n),空间是栈大小,O(logn)。

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
/**
 * 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 {
    int sum;
    public int sumNumbers(TreeNode root) {
        sum = 0;
        dfs(root, 0);
        return sum;
    }
    private void dfs(TreeNode node, int tempSum) {
        if (node == null) {
            return;
        }
        tempSum += node.val;
        if (node.left == null && node.right == null) {
            sum += tempSum;
        }
        tempSum *= 10;
        dfs(node.left, tempSum);
        dfs(node.right, tempSum);
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
class Solution {
   public int sumNumbers(TreeNode root) {
      return dfs(root, 0);
   }
   private int dfs(TreeNode root, int sum) {
      if (root == null) {
         return 0;
      }
      if (root.left == null && root.right == null) {
         return sum * 10 + root.val;
      }
      return dfs(root.left, sum * 10 + root.val) + dfs(root.right, sum * 10 + root.val);
   }
}