leetcode--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.

题意：已经二叉树的中序遍历和后续遍历，重构二叉树

分类：二叉树

解法1：一般重构二叉树的问题，我们都使用递归，然后传入的参数是start，end来标记数组的一段。

这个问题也不例外。

先从后序中找到最后一个节点，就是根节点

然后在中序找到根节点，根节点将中序分成两部分，其实就是左右子树的中序遍历

再再后序从后向前找相同的数目，这个节点把后序分成两部分，其实就是左右子树的后序遍历

递归重复上述过程

递归出口是数组只有一个元素，也就是start==end

/** * 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) {return hepler(postorder, 0, postorder.length-1, inorder, 0, inorder.length-1);    }public TreeNode hepler(int[] postorder,int pstart,int pend,int[] inorder,int istart,int iend){    if(pstart>pend) return null;if(pstart==pend) return new TreeNode(postorder[pstart]);int f = postorder[pend];TreeNode t = new TreeNode(f);int mid = istart;for(;mid<=iend;mid++){if(f==inorder[mid]) break;}int len = mid-istart;t.left = hepler(postorder, pstart, pstart+len-1, inorder, istart, mid-1);t.right = hepler(postorder, pstart+len, pend-1, inorder, mid+1, iend);return t;}}