Leetcode 226 Invert Binary tree flips two fork trees

Daniel does not have a problem to do, we should do a good job

Invert a binary tree.

     4   /     2     7/\   /1   3 6   9

To

     4   /     7     2/\   /9   6 3   1

Trivia:
This problem is inspired by this original tweets by Max Howell:

google:90% of our engineers with the software you wrote (Homebrew), but can ' t invert a binary tree on a Whitebo ard so fuck off.

Recursive solutions:

/** * Definition for a binary tree node. * struct TreeNode {*     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode (int x): Val (x), left (NULL) , right (NULL) {}}; */class Solution {public:    treenode* inverttree (treenode* root)     {        if (root ==null) return root;        treenode* node = inverttree (root->left);        Root->left = Inverttree (root->right);        root->right = node;        return root;    }};

Non-recursive solutions:

Treenode* inverttree (treenode* root)     {        if (root = null) return null;        vector<treenode*> Stack;        Stack.push_back (root);        while (!stack.empty ())        {            treenode* node = Stack.back ();            Stack.pop_back ();            Swap (node->left,node->right);            if (node->left) stack.push_back (node->left);            if (node->right) stack.push_back (node->right);        }        return root;    }

Python:

def inverttree (self, root):    if root:        root.left, root.right = Self.inverttree (root.right), Self.inverttree ( Root.left)        return rootmaybe make it four lines for better Readability:def inverttree (self, root):    if root:        in Vert = Self.inverttree        root.left, root.right = Invert (root.right), invert (root.left)        return Root--------------------------------------------------------------------------------and an iterative version using My own Stack:def inverttree (self, root):    stack = [Root] while    stack:        node = stack.pop ()        if node:            Node.left, node.right = node.right, node.left            stack + = Node.left, node.right    return root

def inverttree (self, root):    If Root is None:        return None    root.left, root.right = Self.inverttree (root.right ), Self.inverttree (Root.left)    return root

Python non-recursive solution:

DFS version:def Inverttree (self, root):        if (root):            self.inverttree (root.left)            Self.inverttree ( Root.right)            root.left, root.right = root.right, root.left            return root   BFS version:def bfs_inverttree (self , root):        queue = Collections.deque ()        if (root):            queue.append (Root) while        (queue):            node = Queue.popleft ()            if (node.left):                queue.append (Node.left)            if (node.right):                queue.append ( Node.right)            node.left, node.right = node.right, node.left        return root

Leetcode 226 Invert Binary tree flips two fork trees