先写了一个最原始的方法1:(java没有切片很难受啊)
/** * 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) { //说明只有右子树 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; } } }
效率低的令人发指,不过想想也是,每次都要重新建立数组,肯定麻烦啊,于是优化,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,可依旧不在第一梯队。