算法系列——Path Sum III

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);
}
}