572. Subtree of Another Tree

最新推荐文章于 2024-09-25 18:14:06 发布

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本文提供了一种高效的算法来解决LeetCode上的一道题：检查一棵树是否是另一棵树的子树。该算法通过递归方式比较两棵树的结构和节点值，实现速度快且内存使用效率高。

题目描述

Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node’s descendants. The tree s could also be considered as a subtree of itself.
在这里插入图片描述

题目链接

https://leetcode.com/problems/subtree-of-another-tree/

方法思路

public class Solution {
    //Runtime: 10 ms, faster than 94.11% 
    //Memory Usage: 41.5 MB, less than 82.09%
    public boolean isSubtree(TreeNode s, TreeNode t) {
        if (s == null) return false;
        if (isSame(s, t)) return true;
        return isSubtree(s.left, t) || isSubtree(s.right, t);
    }
    
    private boolean isSame(TreeNode s, TreeNode t) {
        if (s == null && t == null) return true;
        if (s == null || t == null) return false;
        if (s.val != t.val) return false;
        return isSame(s.left, t.left) && isSame(s.right, t.right);
    }
}

错误的思路：（我不知道错在哪了，可能是想的太复杂了吧）

class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        if(t == null) return true;
        if(s == null && t != null) return false;
        TreeNode node = find(s, t.val);
        List<Integer> ls = inorder(s);
        List<Integer> lt = inorder(t);
        if(ls.size() != lt.size()) return false;
        for(int i = 0; i < ls.size(); i++){
            if(ls.get(i) != lt.get(i))
                return false;
        }
        return true;
    }
    public List<Integer> inorder(TreeNode root){
        List<Integer> ans = new LinkedList<>();
        if(root == null) return ans;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        while(cur != null || !stack.isEmpty()){
            while(cur != null){
                stack.push(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            ans.add(cur.val);
            cur = cur.right;
        }
        return ans;
    }
    public TreeNode find(TreeNode root ,int val){
        if(root == null) return root;
        if(root.val == val) return root;
        TreeNode left = find(root.left, val);
        TreeNode right = find(root.right, val);
        if(left != null) return left;
        if(right != null) return right;
        return null;
    }
}