LeetCode 437. Path Sum III

Posted 2020-09-01 Black_Knight

437. Path Sum III

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• Total Accepted: 18367
• Total Submissions: 47289
• Difficulty: Easy
• Contributors: Stomach_ache

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /      5   -3
   / \      3   2   11
 / \   3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

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 【题目分析】

给定一颗二叉树，找到二叉树中从上往下所有的和值为目标值的节点序列。不要求必须从根节点开始到叶子结点。

【思路】

我们可以遍历二叉树的每一个结点，搜索从这个节点出发的所有满足条件的路径的数目。

【java代码】

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public int pathSum(TreeNode root, int sum) {
12         if(root == null) return 0;
13         int res = pathNum(root, sum);
14         
15         if(root.left != null) res += pathSum(root.left, sum);
16         if(root.right != null) res += pathSum(root.right, sum);
17         
18         return res;
19     }
20     
21     public int pathNum(TreeNode root, int sum) {
22         if(root == null) return 0;
23         int res = 0;
24         if(root.val == sum) res++;
25         if(root.left != null) res += pathNum(root.left, sum-root.val);
26         if(root.right != null) res += pathNum(root.right, sum-root.val);
27         
28         return res;
29     }
30 }