106. Construct Binary Tree from Inorder and Postorder Traversal（由中序遍历和后序遍历构建二叉树）

题目描述

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

Note:
You may assume that duplicates do not exist in the tree.

For example, given

方法思路

由中序遍历和后序遍历确定二叉树
0.确定递归返回的临界条件。
if (post_end < 0 || in_start > in_end) return null;
1.根据后序遍历确定根结点：TreeNode root = new TreeNode(postorder[post_end]);
因为先序遍历的特性，可将preIndex设置为全局变量，很方便。
而后序遍历与中序遍历的组合就无法使用全局变量了，本题只能将post_end 作为参数在递归中进行调用和赋值。
2.在中序遍历的数组中确定根结点的索引：
int index = 0;
for (int i = in_start; i <= in_end; i++)
if (inorder[i] == root.val)
index = i;
3.确定左子树与右子树在中序遍历中的索引范围。
root.left =
build(inorder, postorder, post_end - (in_end - index + 1), in_start, index-1);
root.right = build(inorder, postorder, post_end - 1, index + 1, in_end);
4.return root;

与105的方法相比，本题所使用的方法更具有一般性，可以用来解决105的问题。只需要将post_end改为pre_start即可。

class Solution {
    //Runtime: 7 ms, faster than 46.94%
    //Memory Usage: 37.3 MB, less than 77.17% 
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        return build(inorder, postorder, postorder.length-1, 0, inorder.length-1);
    }
    
    public TreeNode build
    	(int[] inorder, int[] postorder, int post_end, int in_start, int in_end) {
        if (post_end < 0 || in_start > in_end) return null;
        
        TreeNode root = new TreeNode(postorder[post_end]);
				
        int index = 0;
        for (int i = in_start; i <= in_end; i++) 
	        if (inorder[i] == root.val) 
                index = i;

        root.left = 
        	build(inorder, postorder, post_end - (in_end - index + 1), in_start, index-1);
        root.right = build(inorder, postorder, post_end - 1, index + 1, in_end);
     
        return root;
    }
}