数据结构与算法-二叉树-从中序与后序遍历序列构造二叉树

从中序与后序遍历序列构造二叉树

给定两个整数数组 inorder 和 postorder ，其中 inorder 是二叉树的中序遍历， postorder 是同一棵树的后序遍历，请你构造并返回这颗 二叉树 。

示例 1:

输入：inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
输出：[3,9,20,null,null,15,7]

思路：与从前序与中序遍历构造二叉树基本一直，只不过后序遍历中的头节点在最后面

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder == null || postorder == null || inorder.length == 0 || postorder.length == 0) {
            return null;
        }
        int rootVal = postorder[postorder.length - 1];
        TreeNode root = new TreeNode(rootVal);
        if(postorder.length == 1) {
            return root;
        }

        int index = 0;
        for (int i = 0; i < inorder.length; i++) {
            if(inorder[i] == rootVal) {
                index = i;
                break;
            }
        }
        root.left = buildTree(Arrays.copyOfRange(inorder,0,index),Arrays.copyOfRange(postorder,0,index));
        root.right = buildTree(Arrays.copyOfRange(inorder,index+1,inorder.length),Arrays.copyOfRange(postorder,index,postorder.length-1));

        return root;
    }
}