第十三周：113. 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] ] 这道题跟前一题的区别是，返回的是路径，用DFS就可以求解了。 DFS

AC:

public class Solution { private List<List<Integer>> resultList = new ArrayList<List<Integer>>(); public void pathSumInner(TreeNode root, int sum, Stack<Integer>path) { path.push(root.val); if(root.left == null && root.right == null) if(sum == root.val) resultList.add(new ArrayList<Integer>(path)); if(root.left!=null) pathSumInner(root.left, sum-root.val, path); if(root.right!=null)pathSumInner(root.right, sum-root.val, path); path.pop(); } public List<List<Integer>> pathSum(TreeNode root, int sum) { if(root==null) return resultList; Stack<Integer> path = new Stack<Integer>(); pathSumInner(root, sum, path); return resultList; } }