Leetcode105 Construct Binary Tree from Preorder and Inorder Traversal Java實現
阿新 • • 發佈:2018-07-15
提高 bre turn ava mage rsa shm [] img
先寫了一個最原始的方法1:(java沒有切片很難受啊)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public TreeNode buildTree(int[] preorder, int[] inorder) { int len = inorder.length;if(len==0) return null; TreeNode p = new TreeNode(preorder[0]); if(len==1) return p; int index=0;for(;index<len;index++) {if(inorder[index]==preorder[0]) break;} if(index==0) { //說明只有右子樹 int[] newPre = new int[len-1];int[] newIn = new int[len-1];for(int i=0;i<len-1;i++) { newPre[i] = preorder[i+1];newIn[i]=inorder[i+1]; } p.right = buildTree(newPre,newIn); return p; } else if(index==len-1) { int[] newPre = new int[len-1];int[] newIn = new int[len-1];for(int i=0;i<len-1;i++) { newPre[i] = preorder[i+1];newIn[i]=inorder[i]; } p.left = buildTree(newPre,newIn); return p; } 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(int i=0;i<index;i++) { lp[i] = preorder[i+1];li[i]=inorder[i]; } for(int i=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); return p; } } }
效率低的令人發指,不過想想也是,每次都要重新建立數組,肯定麻煩啊,於是優化,得到方法2:
class Solution { public TreeNode buildTree(int[] preorder,int[] inorder) { return build(preorder,inorder,0,preorder.length-1,0,inorder.length-1); } public TreeNode build(int[] preorder,int[] inorder,int pl,int pr,int il,int ir) { int len = pr-pl+1; if(len==0) return null; TreeNode p = new TreeNode(preorder[pl]); if(len==1) return p; int index=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); return p; } }
效果好了一些,可依舊不在第一梯隊。
怎麽優化呢?查看大神思路發現,每次都要在inorder序列中搜索一遍,效率較低,可以建立一個Map存儲inorder中值與index的對應關系,提高效率:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public TreeNode buildTree(int[] preorder,int[] inorder) { Map<Integer,Integer> indexMap = new HashMap<>(); for(int i=0;i<inorder.length;i++) indexMap.put(inorder[i],i); return buildv2(preorder,inorder,0,preorder.length-1,0,inorder.length-1,indexMap); } public TreeNode buildv2(int[] preorder,int[] inorder,int pl,int pr,int il,int ir,Map<Integer,Integer> indexMap) { int len = pr-pl+1; if(len==0) return null; TreeNode p = new TreeNode(preorder[pl]); if(len==1) return p; int index=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); return p; } }
進入第一梯隊,1ms大佬的解法沒看太明白,留待以後再看吧。
Leetcode105 Construct Binary Tree from Preorder and Inorder Traversal Java實現