LeetCode #100 Same Tree

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Problem

Given two binary trees, write a function to check if they are the same or not.

Two binary trees are considered the same if they are structurally identical and the nodes have the same value.

Examples

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Input:     1         1
          / \       / \
         2   3     2   3

        [1,2,3],   [1,2,3]

Output: true
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Input:     1         1
          /           \
         2             2

        [1,2],     [1,null,2]

Output: false
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Input:     1         1
          / \       / \
         2   1     1   2

        [1,2,1],   [1,1,2]

Output: false

Solution1

Method: Recursive Depth First Search

Time Complexity:

Space Complexity:

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

class Solution:
    def isSameTree(self, p, q):
        """
        :type p: TreeNode
        :type q: TreeNode
        :rtype: bool
        """
        if p and q:
            return p.val == q.val and self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)
        else:
            return p == q

Solution2

Method: Iterative Depth First Search

Time Complexity:

Space Complexity:

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

class Solution:
    def isSameTree(self, p, q):
        """
        :type p: TreeNode
        :type q: TreeNode
        :rtype: bool
        """
        if not p and not q:
            return True
        elif not p or not q:
            return False

        stack = [(p, q)]
        
        while stack:
            node_p, node_q = stack.pop()
            if node_p.val != node_q.val:
                return False
            if node_p.left and node_q.left:
                stack.append((node_p.left, node_q.left))
            if node_p.right and node_q.right:
                stack.append((node_p.right, node_q.right))
            if (not node_p.left and node_q.left) or \
               (node_p.left and not node_q.left) or \
               (not node_p.right and node_q.right) or \
               (node_p.right and not node_q.right):
                return False
        
        return True