算法系列——Path Sum III
阿新 • • 發佈:2019-02-14
題目描述
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
解題思路
這道題可以借用之前Path Sum的思想,不過它需要兩層深度優先遍歷。
先從頭結點開始遍歷,求得滿足條件的路徑的個數後,再分別遍歷其左右子樹,求得滿足條件的路徑再累加。
第一種方法:用佇列層次遍歷樹的每個節點,再從每個節點開始dfs求出滿足條件的路徑,將所有的累加就得到結果。
程式實現
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);
}
}