0572. Subtree of Another Tree

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

Description

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

ac

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        return isSub(s, t, true);
    }

    public boolean isSub(TreeNode s, TreeNode t, boolean tIsRoot) {
        // exit
        if (s == null && t == null) return true;
        else if (s == null || t == null) return false;

        return s.val == t.val && isSub(s.left, t.left, false) && isSub(s.right, t.right, false)
            || tIsRoot && isSub(s.left, t, true)
            || tIsRoot && isSub(s.right, t, true);
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        // edge cases
        if (s == null && t != null) return false;

        if (isRootSame(s, t)) return true;

        return isSubtree(s.left, t) || isSubtree(s.right, t);
    }

    private boolean isRootSame(TreeNode s, TreeNode t) {
        if (s == null && t == null) return true;
        else if (s == null || t == null) return false;

        if (s.val != t.val) return false;

        return isRootSame(s.left, t.left) && isRootSame(s.right, t.right);
    }
}

/*
if either return true: 1) root same -> left&&right same; 2) root different, goto children.
*/