📃 题目描述

题目链接：572. 另一棵树的子树 (opens new window)

🔔 解题思路

双递归：判断子树 + 判断两棵树是否相等

和 剑指 Offer 28. 对称的二叉树 (opens new window) 基本差不多

/**
 * 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 {
    // 判断 subRoot 是否是 root 的子树
    public boolean isSubtree(TreeNode root, TreeNode subRoot) {
        if (subRoot == null) {
            return true;
        }
        if (root == null) {
            return false;
        }

        return isSameTree(root, subRoot) || isSubtree(root.left, subRoot) || isSubtree(root.right, subRoot);
    }

    // 判断 root1 和 root2 是否相等
    private boolean isSameTree(TreeNode root1, TreeNode root2) {
        if (root1 == null && root2 == null) {
            return true;
        }
        else if (root1 == null || root2 == null) {
            return false;
        }
        else if (root1.val != root2.val) {
            return false;
        }

        // root1.val == root2.val
        return isSameTree(root1.left, root2.left) && isSameTree(root1.right, root2.right); 
    }
}

💥 复杂度分析

• 空间复杂度：
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