Image of algorithm Problem 22 tree

Source: Internet
Author: User

Topic

Complete a function, enter a binary tree, build its mirrored binary tree

Analysis

The most straightforward solution to this problem is recursion, exchanging left and right subtrees (i.e., exchanging children around).

Code

`1 voidMirrortree (treenode*root)2 {3     if(!root)4         ThrowStd::exception ("Invalid input.");5 6     //swap The left and right subtree7treenode* tmp=root->Pleft;8Root->pleft=root->Pright;9root->pright=tmp;Ten  One     //recursion A     if(root->pleft) -Mirrortree (root->pleft); -     if(root->pright) theMirrortree (root->pright); -  -}`

can also be used in a non-recursive way to solve, for reference to the idea of layered traversal, with a two-way queue, layered traversal of the binary tree, in the process of the traversal of each node of the child can be exchanged

`1 voidMirrorTree1 (treenode*root)2 {3     if(!root)4         ThrowStd::exception ("Invalid input.");5 6Deque<treenode*>dq;7treenode*node;8 Dq.push_back (root);9      while(Dq.size () >0)Ten     { OneNode=Dq.front (); A          -         if(node->pleft==null&&node->pright==NULL) -         { the Dq.pop_front (); -             Continue; -         } -treenode* tmp=node->Pleft; +Node->pleft=node->Pright; -node->pright=tmp; +  A         if(node->pleft) atDq.push_back (node->pleft); -         if(node->pright) -Dq.push_back (node->pright); - Dq.pop_front (); -     } -}`

Image of algorithm Problem 22 tree

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