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] ]
详细思路和代码例如以下:
/** * 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 * 总体思想是遍历。然后加入list逐一试探 * 符合要求的加入结果集 * 不符合要求的删除,然后回溯 */ 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); //加入结果一定要又一次生成实例 list.add(new ArrayList<Integer>(al)); al.remove(al.size()-1);//删除 return; } } al.add(root.val); path(root.left,sum - root.val,al); path(root.right,sum - root.val,al); al.remove(al.size()-1);//一定要删除,确保回溯准确 } }