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 sand 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
题目描述:判断t是不是s的子树
方法一:递归
时间复杂度:o(n) 空间复杂度:o(1)
/** * 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) return false;
// 考虑几种情况,第一种s=t,调用isSubtreeStartRoot(s, t)返回true。第二种情况在根的左子树或右子树上,这需要递归了,因为不知道具体从哪个点开始。 return isSubtreeStartRoot(s, t)||isSubtree(s.left,t)||isSubtree(s.right,t); } private boolean isSubtreeStartRoot(TreeNode s, TreeNode t){ if(s==null&&t==null) return true; //两棵树都走完了,说明两棵树一样,返回true if(s==null||t==null) return false; //一颗树走完了,另一颗没走完,那说明这两个不一样,返回false。 if(s.val!=t.val) return false; return isSubtreeStartRoot(s.left,t.left)&&isSubtreeStartRoot(s.right,t.right); } }