leetCode 106.Construct Binary Tree from Inorder and Postorder Traversal (根据中序遍历和后序遍历构造二叉树)

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：这题和上题类似，前序第一个是根节点，后序遍历最后一个是根节点。其余步骤类似。

代码如下：

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode buildTree(int[] inorder, int[] postorder) {         /**         * 1.根据后序遍历，先确定根节点         * 2.然后在中序遍历中查找根节点，确定根节点在中序遍历的位置         * 3.根据索引位置分割左右子树的前序和中序遍历         * 4.递归求解根节点的左右子树         */         if(postorder.length == 0 || inorder.length == 0)            return null;        TreeNode root = new TreeNode(postorder[postorder.length-1]);        int k = 0;        for(; k < inorder.length; k++){            if(inorder[k] == postorder[postorder.length-1]){                break;            }        }        int[] p1 = Arrays.copyOfRange(postorder,0,k);        int[] q1 = Arrays.copyOfRange(postorder,k,postorder.length-1);        int[] p2 = Arrays.copyOfRange(inorder,0,k);        int[] q2 = Arrays.copyOfRange(inorder,k+1,inorder.length);                root.left = buildTree(p2,p1);        root.right = buildTree(q2,q1);        return root;    }}