572. Subtree of Another Tree

 

 

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.

 

  思路： 

1  写一个判断树是否相同的函数

2 递归的判断当前树节点是否相同。

 

 

 

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:

    def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
        if p == None and q == None:
            return True
        elif p == None and q:
            return False
        elif p and q == None:
            return False
        else:
            if p.val != q.val:
                return False
            return self.isSameTree(p.left,q.left) and self.isSameTree(p.right,q.right)
    def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool: 
        if root == None:
            return False
        if self.isSameTree(root,subRoot):
            return True
        return self.isSubtree(root.left,subRoot) or self.isSubtree(root.right,subRoot)
       
        

 

 

 

class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        if(s == null) return false;
        if(isSame(s,t)) return true;
        return isSubtree(s.left,t) || isSubtree(s.right,t);
    }
    public boolean isSame(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 isSame(s.left,t.left)  && isSame(s.right,t.right); 
        
    }
}