Simple Python solution with modified DFS.

  • 1

    Simple Python solution with modified DFS. Please refer to the comments in the code.

    def dfs2(p1,p2):
        Two pointers.
        p1 searches using DFS, Left-Root-Right manner.
        p2 searches using DFS, Right-Root-Left manner.
        # Base case if both nodes are leaves
        if p1 == None and p2 == None:
            return True
        # Recurse if both of them are not null
        if p1!=None and p2!=None:
            # values for p1 and p2 must be same.
            if p1.val != p2.val:
                return False
            x = dfs2(p1.left,p2.right)
            y = dfs2(p1.right,p2.left)
            # Left of p1 is not same as right of p2 then false.
            # Simply perform "and" operation on x and y to enforce that. 
            return x and y
        # If only one of them is null then return False.
        return False
    class Solution(object):
        def isSymmetric(self, root):
            :type root: TreeNode
            :rtype: bool
            if root == None:
                return True
                return dfs2(root.left,root.right)

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