[LeetCode] 572. 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].
-104 <= root.val <= 104
-104 <= subRoot.val <= 104

另一个树的子树。

给你两棵二叉树 root 和 subRoot 。检验 root 中是否包含和 subRoot 具有相同结构和节点值的子树。如果存在，返回 true ；否则，返回 false 。
二叉树 tree 的一棵子树包括 tree 的某个节点和这个节点的所有后代节点。tree 也可以看做它自身的一棵子树。
来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/subtree-of-another-tree
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思路

题意是给两个树 root 和 subRoot，判断 subRoot 是否是 root 的子树。如果两者相同，也可以视为 subRoot 是 root 的子树。
这个题的思路跟100题same tree非常像。既然是问 subRoot 是否是 root 的子树，那也就意味着 root 这棵树比 subRoot 要大，或者最多相同，所以就可以按照100题的思路递归比较

• root 的左子树和右子树是否分别和 subRoot 一样
• root 是否和 subRoot 是相同的树

复杂度

时间O(n)
空间O(n)

代码

Java实现

/**
 * 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) {
        // corner case
        if (root == null) {
            return false;
        }
        if (subRoot == null) {
            return true;
        }
        return isSubtree(root.left, subRoot) || isSubtree(root.right, subRoot) || helper(root, subRoot);
    }

    private boolean helper(TreeNode p, TreeNode q) {
        if (p == null && q == null) {
            return true;
        }
        if (p == null || q == null) {
            return false;
        }
        if (p.val != q.val) {
            return false;
        }
        return helper(p.left, q.left) && helper(p.right, q.right);
    }
}

JavaScript实现

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} s
 * @param {TreeNode} t
 * @return {boolean}
 */
var isSubtree = function (s, t) {
	if (s === null) {
		return false;
	}
	if (t === null) {
		return true;
	}
	if (isSameTree(s, t)) {
		return true;
	}
	return isSubtree(s.left, t) || isSubtree(s.right, t) || isSameTree(s, t);
};

var isSameTree = function (s, t) {
	if (s === null && t === null) {
		return true;
	}
	if (s === null || t === null) {
		return false;
	}
	if (s.val != t.val) {
		return false;
	}
	return isSameTree(s.left, t.left) && isSameTree(s.right, t.right);
};

相关题目

100. Same Tree
572. Subtree of Another Tree