Lintcode Easy title: Binary Tree inorder traversal two-in-sequence traversal

Topic:

The middle sequence traversal of a two-fork tree

Give a binary tree, return the sequence traversal

Sample Example

Give a two-fork tree {1,#,2,3} ,

   1         2    /   3

Return [1,3,2] .

challenges

Can you use non-recursive algorithms to do this?

Solving:

Direct source of the program

Java Program:

/*** Definition of TreeNode: * public class TreeNode {* public int val; * Public TreeNode left, right; * p Ublic TreeNode (int val) {* This.val = val; * This.left = This.right = null; *} *}*/ Public classSolution {/**     * @paramroot:the root of binary tree. * @return: Inorder in ArrayList which contains node values. */     PublicArraylist<integer>inordertraversal (TreeNode root) {//Write your code hereArraylist<treenode> p =NewArraylist<treenode>(); ArrayList<Integer> res =NewArraylist<integer>();  while(Root! =NULL|| P.size ()! = 0){               while(Root! =NULL) {p.add (root); Root=Root.left; } Root= P.get (P.size ()-1); P.remove (P.size ()-1);              Res.add (Root.val); Root=Root.right; }          returnRes; }    }

View Code

Total time: 1238 Ms

Python program:

"""Definition of Treenode:class treenode:def __init__ (Self, val): Self.val = Val Self.left, self.right = None, none"""classSolution:"""@param root:the root of binary tree.    @return: Inorder in ArrayList which contains node values. """    definordertraversal (self, root):#Write your code herep =[Root] res=[0] whileRoot is  notNoneorLen (p)! = 1:               whileRoot is  notNone:p.append (root) root=Root.left Root= P[len (p)-1]              delP[len (p)-1] Res.append (root.val) root=root.right N=Len (res)returnRES[1:N]

View Code

Total time: 263 ms

Non-recursive procedures, understanding, but also need people ugly to read more

According to the above inspiration, the recursive procedure is as follows:

Java Program:

/*** Definition of TreeNode: * public class TreeNode {* public int val; * Public TreeNode left, right; * PU Blic TreeNode (int val) {* This.val = val; * This.left = This.right = null; *} *}*/ Public classSolution {/**     * @paramroot:the root of binary tree. * @return: Inorder in ArrayList which contains node values. */     PublicArraylist<integer>inordertraversal (TreeNode root) {//Write your code hereArrayList<Integer> res =NewArraylist<integer>(); Res=Inordertrun (res,root); returnRes; }     PublicArraylist<integer> Inordertrun (arraylist<integer>Res,treenode Root) {        if(Root = =NULL)            returnRes; if(root!=NULL){            if(root.left!=NULL) {res=Inordertrun (Res,root.left);            } res.add (Root.val); if(root.right!=NULL) {res=Inordertrun (res,root.right); }        }        returnRes; }    }

View Code

Total time: 1714 Ms

Python program:

"""Definition of Treenode:class treenode:def __init__ (Self, val): Self.val = Val Self.left, self.right = None, none"""classSolution:"""@param root:the root of binary tree.    @return: Inorder in ArrayList which contains node values. """    definordertraversal (self, root):#Write your code hereres =[] Res=Self.inordertrun (res,root)returnResdefInordertrun (self,res,root):ifroot==None:returnResifroot.left!=None:res=Self.inordertrun (res,root.left) res.append (root.val)ifroot.right!=None:res=Self.inordertrun (res,root.right)returnRes

View Code

Total time: 213 ms

Lintcode Easy title: Binary Tree inorder traversal two-in-sequence traversal