leetcode 113. Path Sum II (路徑和) 解題思路和方法
阿新 • • 發佈:2018-01-29
csdn 節點 consola eno == lin sans 要求 又一
Given the below binary tree and?
Given a binary tree and a sum, find all root-to-leaf paths where each path‘s sum equals the given sum.
For example:Given the below binary tree and?
sum
= 22
,
5 / 4 8 / / 11 13 4 / \ / 7 2 5 1
return
[ [5,4,11,2], [5,8,4,5] ]
詳細思路和代碼例如以下:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { /** * 回溯法求解 * 總體思想是遍歷。然後加入list逐一試探 * 符合要求的加入結果集 * 不符合要求的刪除,然後回溯 */ List<List<Integer>> list = new ArrayList<List<Integer>>(); public List<List<Integer>> pathSum(TreeNode root, int sum) { if(root == null){ return list; } path(root,sum,new ArrayList<Integer>()); return list; } private void path(TreeNode root,int sum, List<Integer> al){ if(root == null){ return; } if(root.val == sum){ if(root.left == null && root.right == null){ al.add(root.val); //加入結果一定要又一次生成實例 list.add(new ArrayList<Integer>(al)); al.remove(al.size()-1);//刪除 return; } } al.add(root.val); path(root.left,sum - root.val,al); path(root.right,sum - root.val,al); al.remove(al.size()-1);//一定要刪除,確保回溯準確 } }
leetcode 113. Path Sum II (路徑和) 解題思路和方法