LeetCode -- Path Sum III分析及实现方法

 更新时间:2017年10月25日 14:32:22   投稿:lqh   我要评论
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LeetCode -- Path Sum III分析及实现方法

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.

给定一个二叉树,遍历过程中收集所有可能路径的和,找出和等于X的路径树。

思路:

设当前节点为root,分别收集左右节点路径和的集合,merge到当前集合中;

将当前节点添加到数组中,构成新的可能路径。

实现代码:

/** 
 * 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); 
 } 
} 

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