路径总和 II

给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。

叶子节点 是指没有子节点的节点。

示例 1：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22

输出：[[5,4,11,2],[5,8,4,5]]

思路：深度优先遍历，递归回溯。这里要用到双端队列Deque

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    List<List<Integer>> res = new ArrayList<>();
    Deque<Integer> path = new LinkedList<>();
    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
            if(root == null) {
                return res;
            }
        dfs(root,targetSum);
        return res;
    }

    public void dfs(TreeNode root, int targetSum) {
        if(root == null ) {
            return;
        }
        path.add(root.val);
        targetSum -=root.val;
        if(root.left == null && root.right ==null && targetSum == 0) {
             res.add(new ArrayList<>(path));
        }
        dfs(root.left,targetSum);
        dfs(root.right,targetSum);

        path.pollLast();
    }
}

第二次写：

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    Deque<Integer> path = new LinkedList<>();
    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
            dfs(root,targetSum);
            return res;
    }

    public void dfs(TreeNode root, int targetSum){
        if(root == null) {
            return ;
        }
        path.add(root.val);
        if(root.left == null && root.right == null ) {
            if(targetSum == root.val) {
                res.add(new ArrayList<>(path));
            }
        }

        dfs(root.left,targetSum-root.val);
        dfs(root.right,targetSum-root.val);

        path.pollLast();

    }
}