Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:

You may assume that duplicates do not exist in the tree.

还是重构树  与上一题相比  先序换成了后序  思路还是一样 代码如下：

public class Solution {
public TreeNode buildTree(int[] inorder, int[] postorder) {
if(postorder.length==0&&inorder.length==0){
return null;
}
TreeNode root=new TreeNode(postorder[postorder.length-1]);
int begindex=0;
for(int i=0;i<inorder.length;i++){
if(inorder[i]==postorder[postorder.length-1]){
begindex=i;
break;
}
}
int leftlen=begindex;
int rightlen=inorder.length-begindex-1;
int[] leftpost=new int[leftlen];
int[] leftino=new int[leftlen];
for(int i=0;i<begindex;i++){
leftpost[i]=postorder[i];
leftino[i]=inorder[i];
}
root.left=buildTree(leftino, leftpost);
int[] rightpost=new int[rightlen];
int[] rightino=new int[rightlen];
for(int i=0;i<rightlen;i++){
rightpost[i]=postorder[i+begindex];
rightino[i]=inorder[i+begindex+1];
}
root.right=buildTree(rightino, rightpost);
return root;
}
}