Path Sum II
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] ]
这道题不是自己A出来的，看了discuss区里面的，不过写的的确漂亮
1 package com.gxf.test; 2 3 import java.util.ArrayList; 4 import java.util.List; 5 import java.util.Stack; 6 7 8 9 class TreeNode { 10 int val; 11 TreeNode left; 12 TreeNode right; 13 TreeNode(int x) { val = x; } 14 } 15 16 public class Solution { 17 public List<List<Integer>> pathSum(TreeNode root, int sum) { 18 List<List<Integer>> result = new ArrayList<List<Integer>>(); 19 if(null == root) 20 return result; 21 if(null == root.left && null == root.right){ //叶子结点 22 if(root.val == sum){ 23 List<Integer> element = new ArrayList<Integer>(); 24 element.add(sum); 25 result.add(element); 26 } 27 }else{ 28 result = pathSum(root.left, sum - root.val); //获取左子树所有的路径 29 List<List<Integer>> result_right = pathSum(root.right, sum - root.val);//获取右子树所有路径 30 result.addAll(result_right); 31 32 for(List<Integer> element : result){ 33 element.add(0, root.val); //将根节点添加到第一个位置 34 } 35 } 36 37 return result; 38 } 39 }
另一种DFS
1 package com.gxf.test; 2 3 import java.util.ArrayList; 4 import java.util.List; 5 import java.util.Stack; 6 7 8 9 class TreeNode { 10 int val; 11 TreeNode left; 12 TreeNode right; 13 TreeNode(int x) { val = x; } 14 } 15 16 public class Solution { 17 public List<List<Integer>> pathSum(TreeNode root, int sum) { 18 List<List<Integer>> result = new ArrayList<List<Integer>>(); 19 List<Integer> currentElement = new ArrayList<Integer>(); 20 21 getPath(root, sum, result, currentElement); 22 23 return result; 24 } 25 26 /** 27 * 递归获取result 28 * @param root 29 * @param sum 30 * @param reuslt 31 * @param currentResult 32 */ 33 public void getPath(TreeNode root, int sum, List<List<Integer>> result, List<Integer> currentResult){ 34 if(null == root) 35 return; 36 currentResult.add(root.val); //添加到currentResult 37 if(null == root.left && null == root.right){ //叶子结点 38 if(sum == root.val){ 39 result.add(new ArrayList<Integer>(currentResult)); //添加到result中 40 } 41 currentResult.remove(currentResult.size() - 1); //移除叶子结点 42 return ; 43 } 44 else{ //非叶子结点 45 getPath(root.left, sum - root.val, result, currentResult); //遍历左子树 46 getPath(root.right, sum - root.val, result, currentResult); //遍历右子树 47 48 } 49 50 currentResult.remove(currentResult.size() - 1); //递归回退时，没有这个只会删除叶子结点 51 } 52 }