572. Subtree of Another Tree.md

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

  /**
   * 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

   /**
    * 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;
   	}
   }

Amazon session

• Method 1: recursion

   /**
    * 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 true;
   		return false;
   	}
   	private boolean same(TreeNode s, TreeNode t){
   		if(s == null && t == null) return true;
   		else if((s == null && t != null) || (s != null && t == null)) return false;
   		else{
   			if(s.val != t.val) return false;
   			return same(s.left, t.left) && same(s.right, t.right);
   		}
   	}
   }