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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] ]
相对于普通的dfs有个增添，不满足要求返回时有个删除的操作。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<List<Integer>> pathSum(TreeNode root, int sum) { List<List<Integer>> ans=new ArrayList<List<Integer>>(); if(root==null) return ans; List<Integer> temp = new ArrayList<Integer>(); dfs(root,ans,temp,sum); return ans; } public void dfs(TreeNode root,List<List<Integer>> ans,List<Integer> temp,int sum){ if(root.left == null && root.right == null) { temp.add(root.val); int sum1 = 0; for(int i = 0; i < temp.size(); i++) { sum1 += temp.get(i); } if(sum1 == sum) ans.add(new ArrayList<Integer>(temp)); return ; } temp.add(root.val); if(root.left != null) { dfs(root.left, ans, temp, sum); temp.remove(temp.size() - 1); } if(root.right != null) { dfs(root.right, ans, temp, sum); temp.remove(temp.size() - 1); } } }
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原文地址：http://www.cnblogs.com/gonewithgt/p/4584613.html
