Path Sum II 解答

Question

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]
]

Solution

Traditional way, use DFS and recursion.

 /**
  * 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>> result = new ArrayList<List<Integer>>();
         List<Integer> prevList = new ArrayList<Integer>();
         dfs(root, sum, result, prevList);
         return result;
     }

     private void dfs(TreeNode root, int target, List<List<Integer>> result, List<Integer> prevList) {
         if (root == null)
             return;
         prevList.add(root.val);

         if (root.left == null && root.right == null) {
             if (root.val == target)
                 result.add(new ArrayList<Integer>(prevList));
         } else {
             List<Integer> tmpList2 = new ArrayList<Integer>(prevList);
             if (root.left != null)
                 dfs(root.left, target - root.val, result, prevList);
             if (root.right != null)
                 dfs(root.right, target - root.val, result, tmpList2);
         }

     }
 }