给你二叉树的根节点 root 和一个表示目标和的整数 targetSum 。判断该树中是否存在 根节点到叶子节点 的路径，这条路径上所有节点值相加等于目标和 targetSum 。如果存在，返回 true ；否则，返回 false 。

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,null,1], targetSum = 22
输出：true
解释：等于目标和的根节点到叶节点路径如上图所示。

/**
 * 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 {
    public boolean hasPathSum(TreeNode root, int targetSum) {
        // 如果根节点为空，则返回false
        if (root == null) return false;
        
        // 如果当前节点是叶子节点，且节点值等于目标和，则返回true
        if (root.left == null && root.right == null && targetSum == root.val) {
            return true;
        }
        
        // 递归检查左子树
        boolean leftPathSum = false;
        if (root.left != null) {
            leftPathSum = hasPathSum(root.left, targetSum - root.val);
        }
        
        // 递归检查右子树
        boolean rightPathSum = false;
        if (root.right != null) {
            rightPathSum = hasPathSum(root.right, targetSum - root.val);
        }
        
        // 返回左右子树的结果
        return leftPathSum || rightPathSum;
    }
}

回溯更好的展现出来 

/**
 * 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 {
    public boolean tral(TreeNode root,int count){
        if(root==null) return false;
        if(root.left==null&&root.right==null&&count==0) return true;
        if(root.left!=null){
            count-=root.left.val;
            if(tral(root.left,count)) return true;
            count+=root.left.val;//回溯
        }
         if(root.right!=null){
            count-=root.right.val;
            if(tral(root.right,count)) return true;
            count+=root.right.val;//回溯
        }
        return false;
    }
    
    public boolean hasPathSum(TreeNode root, int targetSum) {
        if(root==null) return false;
        return tral(root,targetSum-root.val);
    }
}

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]]

提示：

• 树中节点总数在范围 [0, 5000] 内
• -1000 <= Node.val <= 1000
• -1000 <= targetSum <= 1000

/**
 * 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 {
    public void travelsal(TreeNode root,int count,List<List<Integer>> result,List<Integer> path){
        if(root==null) return;
        path.add(root.val);//这里将遍历的节点值添加到path中
        if(root.left == null&&root.right==null&&count-root.val==0) result.add(new ArrayList<>(path));
        if(root.left!=null){
            //这里就不用add了
            travelsal(root.left,count-root.val,result,path);
        }
        if(root.right!=null){
             //这里就不用add了
            travelsal(root.right,count-root.val,result,path);
        }
        path.remove(path.size()-1);//回溯，移除当前节点，恢复路径状态!!!!!!
    }
    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
        List<List<Integer>> result = new ArrayList<>();
        if(root==null) return result;
        List<Integer> path = new ArrayList<>();
         travelsal(root,targetSum,result,path);
        return result;
    }
}