golang 解法, 执行用时: 4 ms, 内存消耗: 6.7 MB, 提交时间: 2023-10-25 23:04:34

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */

func isSubPath(head *ListNode, root *TreeNode) bool {
	if root == nil {
		return false
	}
	return isSubPathHelper(head, root) || isSubPath(head, root.Left) || isSubPath(head, root.Right)
}

func isSubPathHelper(head *ListNode, root *TreeNode) bool {
	if head == nil {
		return true
	}
	if root == nil {
		return false
	}
	if head.Val != root.Val {
		return false
	}
	return isSubPathHelper(head.Next, root.Right) || isSubPathHelper(head.Next, root.Left)
}

python3 解法, 执行用时: 88 ms, 内存消耗: 18.2 MB, 提交时间: 2023-10-25 23:04:02

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def dfs(self, head: ListNode, rt: TreeNode) -> bool:
        if not head:
            return True
        if not rt:
            return False
        if rt.val != head.val:
            return False
        return self.dfs(head.next, rt.left) or self.dfs(head.next, rt.right)

    def isSubPath(self, head: ListNode, root: TreeNode) -> bool:
        if not root:
            return False
        return self.dfs(head, root) or self.isSubPath(head, root.left) or self.isSubPath(head, root.right)

javascript 解法, 执行用时: 76 ms, 内存消耗: 47.7 MB, 提交时间: 2023-10-25 23:03:38

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {ListNode} head
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSubPath = function(head, root) {
    if (root == null) return 0;
    return dfs(root, head) || isSubPath(head, root.left) || isSubPath(head, root.right);
};
var dfs = function(rt, head) {
    if (head == null) return true;
    if (rt == null) return false;
    if (rt.val != head.val) return false;
    return dfs(rt.left, head.next) || dfs(rt.right, head.next);
}

cpp 解法, 执行用时: 28 ms, 内存消耗: 31.8 MB, 提交时间: 2023-10-25 23:03:16

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
    bool dfs(TreeNode* rt, ListNode* head) {
        // 链表已经全部匹配完，匹配成功
        if (head == NULL) return true;
        // 二叉树访问到了空节点，匹配失败
        if (rt == NULL) return false;
        // 当前匹配的二叉树上节点的值与链表节点的值不相等，匹配失败
        if (rt->val != head->val) return false;
        return dfs(rt->left, head->next) || dfs(rt->right, head->next);
    }
public:
    bool isSubPath(ListNode* head, TreeNode* root) {
        if (root == NULL) return false;
        return dfs(root, head) || isSubPath(head, root->left) || isSubPath(head, root->right);
    }
};

cpp 解法, 执行用时: 24 ms, 内存消耗: 31.7 MB, 提交时间: 2023-10-25 23:02:49

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isSubPath(ListNode* head, TreeNode* root) {
        if(head == nullptr) return true;
        if(root == nullptr) return false;
        return dfs(head,root) || isSubPath(head,root->left) || isSubPath(head,root->right);
    }
    bool dfs(ListNode *head,TreeNode *node){
        if(head == nullptr) return true;
        if(node == nullptr) return false;
        if(head->val != node->val) return false;
        return dfs(head->next,node->left) || dfs(head-> next,node->right);
    }
};

java 解法, 执行用时: 1 ms, 内存消耗: 42.3 MB, 提交时间: 2023-10-25 23:02:32

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
/**
 * 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 isSubPath(ListNode head, TreeNode root) {
        if (head == null) {
            return true;
        }
        if (root == null) {
            return false;
        }
        //先判断当前的节点，如果不对，再看左子树和右子树呗
        return isSub(head, root) || isSubPath(head, root.left) || isSubPath(head, root.right);
    }

    private boolean isSub(ListNode head, TreeNode node) {
        //特判：链表走完了，返回true
        if (head == null) {
            return true;
        }
        //特判：链表没走完，树走完了，这肯定不行，返回false
        if (node == null) {
            return false;
        }
        //如果值不同，必定不是啊
        if (head.val != node.val) {
            return false;
        }
        //如果值相同，继续看，左边和右边有一个满足即可
        return isSub(head.next, node.left) || isSub(head.next, node.right);
    }
}