Leetcode105 Construct Binary Tree from preorder and inorder traversal Java implementation

First wrote a primitive Method 1: (Java without slicing very uncomfortable AH)

/*** Definition for a binary tree node. * public class TreeNode {* int val; * TreeNode left; * TreeNode rig Ht * TreeNode (int x) {val = x;} }*/classSolution { PublicTreeNode Buildtree (int[] Preorder,int[] inorder) {        intLen =inorder.length; if(len==0)return NULL; TreeNode P=NewTreeNode (preorder[0]); if(len==1)returnp; intindex=0; for(; index<len;index++) {if(Inorder[index]==preorder[0]) Break;} if(index==0) {//description only right subtree            int[] Newpre =New int[Len-1];int[] Newin =New int[Len-1];  for(inti=0;i<len-1;i++) {Newpre[i]= Preorder[i+1];newin[i]=inorder[i+1]; } p.right=Buildtree (Newpre,newin); returnp; }        Else if(index==len-1) {            int[] Newpre =New int[Len-1];int[] Newin =New int[Len-1];  for(inti=0;i<len-1;i++) {Newpre[i]= preorder[i+1];newin[i]=Inorder[i]; } p.left=Buildtree (Newpre,newin); returnp; }        Else {            int[] LP =New int[index];int[] Li =New int[index]; int[] RP =New int[Len-index-1];int[] ri=New int[len-1-index];  for(inti=0;i<index;i++) {Lp[i]= preorder[i+1];li[i]=Inorder[i]; }             for(inti=index+1;i<len;i++) {Rp[i-index-1]=preorder[i];ri[i-index-1]=Inorder[i]; } p.left=Buildtree (Lp,li); P.right=Buildtree (Rp,ri); returnp; }    }}

Inefficient and heinous, but think also, every time to re-establish the array, certainly trouble ah, so optimization, get Method 2:

classSolution { PublicTreeNode Buildtree (int[] Preorder,int[] inorder) {        returnBuild (preorder,inorder,0,preorder.length-1,0,inorder.length-1); }     PublicTreeNode Build (int[] Preorder,int[] inorder,intPlintprintIlintir) {        intLen = pr-pl+1; if(len==0)return NULL; TreeNode P=NewTreeNode (PREORDER[PL]); if(len==1)returnp; intIndex=il; for(; index<ir+1;index++) {if(INORDER[INDEX]==PREORDER[PL]) Break;} P.left= Build (preorder,inorder,pl+1,pl+index-il,il,index-1); P.right= Build (preorder,inorder,pl+index-il+1,pr,index+1, IR); returnp; }}

The effect is better, but still not the first echelon.

How to optimize it? See the idea of the great God found that each time in the inorder sequence to search again, less efficient, you can establish a map storage inorder the corresponding relationship between the value and index, improve efficiency:

/*** Definition for a binary tree node. * public class TreeNode {* int val; * TreeNode left; * TreeNode rig Ht * TreeNode (int x) {val = x;} }*/classSolution { PublicTreeNode Buildtree (int[] Preorder,int[] inorder) {Map<Integer,Integer> Indexmap =NewHashmap<>();  for(inti=0;i<inorder.length;i++) Indexmap.put (inorder[i],i); returnBuildv2 (preorder,inorder,0,preorder.length-1,0,inorder.length-1, Indexmap); }     PublicTreeNode Buildv2 (int[] Preorder,int[] inorder,intPlintprintIlintIr,map<integer,integer>Indexmap) {        intLen = pr-pl+1; if(len==0)return NULL; TreeNode P=NewTreeNode (PREORDER[PL]); if(len==1)returnp; intindex=Indexmap.get (PREORDER[PL]); P.left= Buildv2 (preorder,inorder,pl+1,pl+index-il,il,index-1, Indexmap); P.right= Buildv2 (preorder,inorder,pl+index-il+1,pr,index+1, Ir,indexmap); returnp; }}

Into the first echelon, 1ms big guy's solution did not see too understand, left to see it later.

Leetcode105 Construct Binary Tree from preorder and inorder traversal Java implementation