572. Subtree of Another Tree
by Botao Xiao
572. Subtree of Another Tree
Question
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.
Example 1:
Given tree s:
3
/ \
4 5
/ \
1 2
Given tree t:
4
/ \
1 2
Return true, because t has the same structure and node values with a subtree of s.
Example 2:
Given tree s:
3
/ \
4 5
/ \
1 2
/
0
Given tree t:
4
/ \
1 2
Return false.
Solution:
- Method 1: Recursion
```Java
/**
- 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) { if(s != null && t != null) return same(s, t) || (isSubtree(s.left, t) || isSubtree(s.right, t)); else if(s != null || t != null) return false; else return true; } private boolean same(TreeNode s, TreeNode t){ if(s != null && t != null) return s.val == t.val && same(s.left, t.left) && same(s.right, t.right); else if(s != null || t != null) return false; else return true; } } ```
Second Time
- Method 1: Recursion
```Java
/**
- 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) { if(s != null && t != null){ return isSame(s, t) || isSubtree(s.left, t) || isSubtree(s.right, t); }else if(s == null && t == null) return true; return false; } private boolean isSame(TreeNode s, TreeNode t){ if(s != null && t != null){ if(s.val != t.val) return false; return isSame(s.left, t.left) && isSame(s.right, t.right); }else if(s == null && t == null) return true; return false; } } ```
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