Leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal 中序和后序恢复二叉树

1 解题思想

2 原题

Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.

3 AC解

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
/**
* post:left-right-root
* inorder:left-root-right
* */
public class Solution {
/**
* 核心思想
* 同上题，只是需要记得postorder从后面开始，然后改一下位置参数就可以了
* */
public TreeNode buildTreeHelper(int[] inorder,int is,int ie,int[] preorder,int ps) {
if(is>ie || ps < 0) return null;
//首先从inorder中确定第一个，然后既可以划分左右子树
//System.out.println(ps);
TreeNode root = new TreeNode(preorder[ps]);
int rootAtIn = is;
while(inorder[rootAtIn] != root.val) rootAtIn ++;
//拆分成两个子问题
root.left =  buildTreeHelper(inorder,is,rootAtIn - 1,preorder,ps-(ie- rootAtIn + 1));
root.right = buildTreeHelper(inorder,rootAtIn+1,ie,preorder,ps-1);
return root;
}

public TreeNode buildTree(int[] inorder, int[] postorder) {
return buildTreeHelper(inorder,0,inorder.length-1,postorder,postorder.length-1);
}
}