标签：
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] ]
理解下leaf node:先序遍历，先加root，然后遍历左右子树，如果都为null（遍历完），回溯。
4
/ \
null null
/** * 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>> res=new ArrayList<>(); List<Integer> check=new ArrayList<Integer>(); dps(root,res,check,sum); return res; } public void dps(TreeNode root, List<List<Integer>> res,List<Integer> check,int sum) { if(root==null) { return; } check.add(root.val); if(root.left==null&&root.right==null&&sum==root.val) { res.add(new ArrayList<>(check)); } else { dps(root.left,res,check,sum-root.val); dps(root.right,res,check,sum-root.val); } check.remove(check.size()-1); } }
/** * 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>> res=new ArrayList<>(); List<Integer> check=new ArrayList<Integer>(); dps(root,res,check,sum); return res; } public void dps(TreeNode root, List<List<Integer>> res,List<Integer> check,int sum) { if(root==null) { return; } if(root.left==null&&root.right==null) { if(sum==root.val) { List<Integer> tmp=new ArrayList<>(check); tmp.add(root.val); res.add(tmp); } return; } check.add(root.val); dps(root.left,res,check,sum-root.val); dps(root.right,res,check,sum-root.val); check.remove(check.size()-1); } }
标签：
原文地址：http://www.cnblogs.com/Machelsky/p/5902332.html
