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);
        }
        
    }
}