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序遍历，求二叉树的结构

思路：通过后序遍历中的最后一个元素，然后通过中序遍历知道左子树的包含元素和右子树的包含元素。根据左子树的中序遍历和后序遍历构造左子树，右子树的中序遍历和后序遍历构造右子树。

代码如下：

//二叉树的结构表示为
class TreeNode
{int val;TreeNode left;TreeNode right;TreeNode(int x) {val = x;}
}
class Solution
{public TreeNode buildTree(int[] inorder, int[] postorder){int len = postorder.length;if (0 == len) return null;TreeNode root = new TreeNode(postorder[len - 1]);int index = 0;for (int i = 0; i < inorder.length; i++){if (inorder[i] == postorder[len - 1]){index = i;break;}}int[] leftinorder = new int[index];int[] leftpostorder = new int[index];int[] rightinorder = new int[len - index - 1];int[] rightpostorder = new int[len - index - 1];for (int i = 0; i < index; i++){leftinorder[i] = inorder[i];leftpostorder[i] = postorder[i];}for (int i = 0; i < len - index - 1; i++){rightinorder[i] = inorder[index + 1 + i];rightpostorder[i] = postorder[index + i];}root.left = buildTree(leftinorder, leftpostorder);root.right = buildTree(rightinorder, rightpostorder);return root;}
}

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