leetcode113- Path Sum II- medium

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]
]

DFS。这题叶子节点和null分开处理。要小心叶子节点那里允许加个别点的时候，加完result也要remove！所有crt buffer里面的add 和 remove操作必须同时搭配！！

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> crt = new ArrayList<>();
        dfs(root, sum, crt, result);
        return result;
    }
    
    private void dfs(TreeNode root, int dist, List<Integer> crt, List<List<Integer>> result) {
        
        if (root == null) {
            return;
        }
        
        if (root.left == null && root.right == null && root.val == dist) {
            crt.add(root.val);
            result.add(new ArrayList<Integer>(crt));
            // 千万小心这里也得remove！！
            crt.remove(crt.size() - 1);
            return;
        }
        
        crt.add(root.val);
        dfs(root.left, dist - root.val, crt, result);
        dfs(root.right, dist - root.val, crt, result);
        crt.remove(crt.size() - 1);
        
    }
}