p437

发表于 | 分类于 |

LeetCode p437 Path Sum III 题解

1.题目：

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

题意：

给一个树和一个目标数，求这个树从任意点开始有多少条路径可以得到目标数。

2.解题思路:

在每一个节点有四种向下的寻找方式，用set储存找过的节点。

3.代码

[title] [] [url] [link text]
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
37
38
39
 
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int target;
	public HashSet<TreeNode> hashSet;

	public int pathSum(TreeNode root, int sum) {

		target = sum;
		hashSet = new HashSet<TreeNode>();
		return pathSum(root, sum, false);
	}

	public int pathSum(TreeNode root, int sum, boolean hasparent) {

		if (root == null)
			return 0;

		if (hashSet.contains(root) && !hasparent)
			return 0;
		if (!hasparent)
			hashSet.add(root);

		int count = sum == root.val ? 1 : 0;
		count += pathSum(root.left, sum - root.val, true);
		count += pathSum(root.right, sum - root.val, true);
		count += pathSum(root.left, target, false);
		count += pathSum(root.right, target, false);
		return count;
	}
}

4.一些总结：