Path Sum III
阿新 • • 發佈:2018-11-17
子節點 path sum you .com tree lee div tps sta
https://leetcode.com/problems/path-sum-iii
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:
這題雖然簡單,寫對了可不容易,想寫出線性的方法更不容易。
簡單的解法,一個個節點看。每個節點都往下找他的所有子節點,路徑和為sum的,就算一個。
這樣做的時間復雜度是O(n^2)。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution {public int pathSum(TreeNode root, int sum) { int[] count = new int[1]; count(root, sum, count); if (root != null) { count[0] += pathSum(root.left, sum); count[0] += pathSum(root.right, sum); } return count[0]; } public void count(TreeNode root, int sum, int[] count) { if (root == null) { return; } if (root.val == sum) { count[0]++; } if (root.left != null) { count(root.left, sum - root.val, count); } if (root.right != null) { count(root.right, sum - root.val, count); } } }
Path Sum III