[LeetCode] 106. Construct Binary Tree from Inorder and Postorder Traversal Java

题目;

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

 题意及分析：给出棵二叉树的中序遍历和后续遍历，求这棵树。同用先序和中序推二叉树，不同的点是，后序遍历的最后一个点才是根节点。

代码：

/**
 * 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) {
        if(inorder.length==0) return null;
        TreeNode root = new TreeNode(postorder[postorder.length-1]);        //后序遍历最后一个节点为根节点
        build(root,inorder,postorder);
        return root;
    }

    public void build(TreeNode root,int[] inorder, int[] postorder){
        int nodeVal = root.val;

        int index = 0;
        for(int i=0;i<inorder.length;i++){
            if(inorder[i]==nodeVal){        //找到中序遍历分割左右子树的点
                index = i;
                break;
            }
        }
        int[] leftInorder=Arrays.copyOfRange(inorder,0,index);      //左子树中序遍历
        int[] leftPostorder=Arrays.copyOfRange(postorder,0,index);    //左子树后序遍历
        if(leftPostorder.length!=0){        //后序遍历的最后一个值为当前点的左节点
            root.left = new TreeNode(leftPostorder[leftPostorder.length-1]);
            build(root.left,leftInorder,leftPostorder);
        }
        
        int[] rightInorder=Arrays.copyOfRange(inorder,index+1,inorder.length);
        int[] rightPostorder=Arrays.copyOfRange(postorder,index,postorder.length-1);
        if(rightInorder.length!=0){
            root.right = new TreeNode(rightPostorder[rightPostorder.length-1]);
            build(root.right,rightInorder,rightPostorder);
        }
    }
}