二叉树中和为某一值的路

本题需要遍历每个路径，首先考虑深度优先搜索

使用栈避免递归，每次到达叶子节点进行判断该路径的和是否为target

最后需要考虑排序，即按list长度降序排序

代码如下

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Stack;
public class Soultion {
    class TreeNode {
        int val = 0;
        TreeNode left = null;
        TreeNode right = null;
        public TreeNode(int val) {
            this.val = val;
        }
    }
    class Com implements Comparator> {
        @Override
        public int compare(ArrayList o1, ArrayList o2) {
            // TODO Auto-generated method stub
            return -1 * (o1.size() - o2.size());//降序排序
        }

    }
    private void def(TreeNode root, int target, ArrayList> res, ArrayList cur) {
        if(root == null) return ;
        Stack stack = new Stack();

        stack.push(root);
        int sum = 0;
        while(!stack.isEmpty()) {
            TreeNode node = stack.pop();
            sum += node.val;
            cur.add(node.val);
            //判断路径的和是否为target，到达叶子节点无论和是否为target，都需要将叶子节点的值减去并且将该节点从路径cur中去掉
            if(sum == target && node.left == null && node.right == null) {
                sum -= node.val;
                res.add(new ArrayList(cur));
                cur.remove(cur.size() - 1);
            } else if(sum != target && node.left == null && node.right == null) { 
                sum -= node.val;
                cur.remove(cur.size() - 1);
            }
            if(node.left != null) {
                stack.push(node.left);
            }
            if(node.right != null) {
                stack.push(node.right);
            }
        }
    }

    public ArrayList> FindPath(TreeNode root,int target) {

        ArrayList> res = new ArrayList();
        ArrayList cur = new ArrayList();
        if(root == null) return res;

        def(root, target, res, cur);

        Collections.sort(res, new Com());

        return res;

    }
}