本文共 818 字，大约阅读时间需要 2 分钟。

思路：根据后序先找出节点，然后去中序找出左右，依次递归。

/** * 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 build(inorder, postorder,0,inorder.length-1,0,postorder.length-1);    }    public TreeNode build(int[] inorder, int[] postorder,int is,int ie,int ps,int pe){    	if(ps>pe){    		return null;    	}    	int val=postorder[pe];    	TreeNode node=new TreeNode(val);    	int i = is;    	for (; i <=ie; i++) {			if (inorder[i]==val) {				break;			}		}    	int rang=i-is;    	node.left=build(inorder, postorder,is,i-1,ps,ps+rang-1);    	node.right=build(inorder,postorder,i+1,ie,ps+rang,pe-1);    	return node;    	    }}

耗时：400ms，中游