算法系列——Path Sum III
2017-08-24 12:29
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题目描述
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
解题思路
这道题可以借用之前Path Sum的思想,不过它需要两层深度优先遍历。先从头结点开始遍历,求得满足条件的路径的个数后,再分别遍历其左右子树,求得满足条件的路径再累加。
第一种方法:用队列层次遍历树的每个节点,再从每个节点开始dfs求出满足条件的路径,将所有的累加就得到结果。
程序实现
public class Solution { public int pathSum(TreeNode root, int sum) { if(root==null) return 0; return dfs(root,sum)+pathSum(root.left,sum)+pathSum(root.right,sum); } private int dfs(TreeNode root,int sum){ if(root==null) return 0; int count=0; if(sum==root.val) count++; return count+dfs(root.left,sum-root.val)+dfs(root.right,sum-root.val); } }
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