Subtree of Another Tree - Tree - Easy - LeetCode
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Subtree of Another Tree - Tree - Easy - LeetCode

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  • 1The problem involves checking if tree t is a subtree of tree s with matching structure and node values.
  • 2A subtree consists of a node in s and all its descendants, including s itself.
  • 3The solution uses a recursive approach to compare nodes and their children, with a time complexity of O(2^n).

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AskGif

"The problem involves checking if tree t is a subtree of tree s with matching structure and node values."

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)

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sumitc91

Published on 7 October 2020 · 1 min read · 245 words

Part of AskGif Blog · coding

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