106. 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.

思路:根据后序遍历的最后一个值为根节点，在inorder中查找到这个节点，左边的为左子树，右边为右子树

代码如下( 已通过leetcode)

public class Solution {
   public TreeNode buildTree(int[] inorder, int[] postorder) {
    if(inorder.length==0) return null;
    TreeNode root=new TreeNode(postorder[postorder.length-1]);
    int pos=0;
    for(int i=0;i<inorder.length;i++) {
    if(inorder[i]==postorder[inorder.length-1]) {
    pos=i;
    break;
    }
    }
   
    int[] leftinorder=new int[pos];
    int[] leftpostorder=new int[pos];
    for(int i=0;i<pos;i++) {
    leftinorder[i]=inorder[i];
    leftpostorder[i]=postorder[i];
    }
   
   
    int[] rightinorder=new int[inorder.length-pos-1];
    int[] rightpostorder=new int[inorder.length-pos-1];
    for(int i=0;i<inorder.length-pos-1;i++) {
    rightinorder[i]=inorder[pos+i+1];
    rightpostorder[i]=postorder[pos+i];
    }
    //System.out.println(root.val);
    root.left=buildTree(leftinorder, leftpostorder);
    root.right=buildTree(rightinorder, rightpostorder);
    return root;
   }
}