2018-05-02 10:59:16 1360瀏覽
廢話不多說了,下面和大家分享一下扣丁學(xué)堂Java在線學(xué)習(xí)課程:LeetCode -- Path Sum III分析及實現(xiàn)方法,希望可以幫到對Java開發(fā)感興趣的小伙伴們。
LeetCode -- Path Sum III分析及實現(xiàn)方法
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.
給定一個二叉樹,遍歷過程中收集所有可能路徑的和,找出和等于X的路徑樹。
思路:
設(shè)當(dāng)前節(jié)點為root,分別收集左右節(jié)點路徑和的集合,merge到當(dāng)前集合中;
將當(dāng)前節(jié)點添加到數(shù)組中,構(gòu)成新的可能路徑。
實現(xiàn)代碼:
/** * Definition for a binary tree node. * public class TreeNode { * public int val; * public TreeNode left; * public TreeNode right; * public TreeNode(int x) { val = x; } * } */ public class Solution { private int _sum; private int _count; public int PathSum(TreeNode root, int sum) { _count = 0; _sum = sum; Travel(root, new List<int>()); return _count; } private void Travel(TreeNode current, List<int> ret){ if(current == null){ return ; } if(current.val == _sum){ _count ++; } var left = new List<int>(); Travel(current.left, left); var right = new List<int>(); Travel(current.right, right); ret.AddRange(left); ret.AddRange(right); for(var i = 0;i < ret.Count; i++){ ret[i] += current.val; if(ret[i] == _sum){ _count ++; } } ret.Add(current.val); //Console.WriteLine(ret); } }
好了,以上就是小編給大家分享的Java在線學(xué)習(xí)課程:LeetCode -- Path Sum III分析及實現(xiàn)方法,想要學(xué)好Java開發(fā)的小伙伴一定要選擇專業(yè)的Java培訓(xùn)機(jī)構(gòu),小編給大家推薦專業(yè)的Java培訓(xùn)機(jī)構(gòu)扣丁學(xué)堂,扣丁學(xué)堂不僅有專業(yè)的老師和與時俱進(jìn)的課程體系還有大量的Java在線教程供學(xué)員觀看學(xué)習(xí),心動的小伙伴快快行動吧。Java技術(shù)交流群:670348138。
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