LeetCode-572. Subtree of Another TreeLevel up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.https://leetcode.com/problems/subtree-of-another-tree/

Given the roots of two binary trees root and subRoot, return true if there is a subtree of root with the same structure and node values of subRoot and false otherwise.

A subtree of a binary tree tree is a tree that consists of a node in tree and all of this node's descendants. The tree tree could also be considered as a subtree of itself.

Example 1:

Input: root = [3,4,5,1,2], subRoot = [4,1,2]
Output: true

Example 2:

Input: root = [3,4,5,1,2,null,null,null,null,0], subRoot = [4,1,2]
Output: false

Constraints:

• The number of nodes in the root tree is in the range [1, 2000].
• The number of nodes in the subRoot tree is in the range [1, 1000].
• -10^4 <= root.val <= 10^4
• -10^4 <= subRoot.val <= 10^4

【C++】

/**
 * 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 isSubtree(TreeNode* root, TreeNode* subRoot) {
        if (root == nullptr) { return false;}
        return isSubtreeWithRoot(root, subRoot)
            || isSubtree(root->left, subRoot)
            || isSubtree(root->right, subRoot);
    }
    bool isSubtreeWithRoot(TreeNode* pRoot1, TreeNode* pRoot2) {
        if (!pRoot1 && !pRoot2) {return true;}
        if (!pRoot1 || !pRoot2) {return false;}
        if (pRoot1->val != pRoot2->val) {return false;}
        return isSubtreeWithRoot(pRoot1->left, pRoot2->left)
            && isSubtreeWithRoot(pRoot1->right, pRoot2->right);
    }
};

【Java】

/**
 * 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 isSubtree(TreeNode root, TreeNode subRoot) {
        if (root == null) {return false;}
        return isSubtreeWithRoot(root, subRoot)
            || isSubtree(root.left, subRoot)
            || isSubtree(root.right, subRoot);
    }
    boolean isSubtreeWithRoot(TreeNode pRoot1, TreeNode pRoot2) {
        if (pRoot1 == null && pRoot2 == null) {return true;}
        if (pRoot1 == null || pRoot2 == null) {return false;}
        if (pRoot1.val != pRoot2.val) {return false;}
        return isSubtreeWithRoot(pRoot1.left, pRoot2.left)
            && isSubtreeWithRoot(pRoot1.right, pRoot2.right);
    }
}