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by Sumit Chourasia | Oct 07, 2020 | Category :coding | Tags : algorithm binary-tree data-structure easy leetcode tree

Subtree of Another Tree - Tree - Easy - LeetCode

Subtree of Another Tree - Tree - Easy - LeetCode

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.
 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public bool IsSubtree(TreeNode s, TreeNode t) {
        if(t==null){
            return true;
        }
        if(s==null && t!=null){
            return false;
        }
        
        if(s.val==t.val){
             if(IsEqual(s.left,t.left) && IsEqual(s.right,t.right)){
                 return true;
             }
        }
        
        return IsSubtree(s.left,t) || IsSubtree(s.right,t);
    }
    
    private bool IsEqual(TreeNode s, TreeNode t){
        if(s==null && t == null){
            return true;
        }
        
        if(s==null || t == null){
            return false;
        }
        
        if(s.val != t.val){
            return false;
        }
        
        return IsEqual(s.left,t.left) && IsEqual(s.right,t.right);
    }
}

Time Complexity: O(2^n)

Space Complexity: O(1)