LeetCode -- Construct Binary Tree from Inorder and Postorder Traversal

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.




就是输入中序遍历（左右中）和后序遍历（左右中）序列，生成二叉树。


思路：


设iFrom, iTo 分别inorder的起始和终止索引； pFrom ,pTo为递归过程中postorder的起始和终止索引。


1. 每次取后序遍历的最后节点，作为当前根节点，即current = new TreeNode(postorder[len - 1])
2. 从inorder序列中找到postorder[len-1]的索引，记为index
3. 创建左子树： inorder的索引范围：[0,index) , postorder的索引范围：[pFrom, pFrom + 距离(index, iFrom) - 1)。
4. 创建右子树： inorder的索引范围: [index + 1, len), postorder 的索引范围: [pFrom + 距离(index, iFrom), pTo-1]。


终止条件：pFrom > pTo 或iFrom > iTo






实现代码：

/** * Definition for a binary tree node. * public class TreeNode { *     public int val; *     public TreeNode left; *     public TreeNode right; *     public TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode BuildTree(int[] inorder, int[] postorder) {TreeNode node = null;var len = postorder.Length - 1;Build(ref node, inorder, 0, len, postorder, 0, len);return node;}private void Build(ref TreeNode current, int [] inorder, int iFrom, int iTo, int[] postorder, int pFrom, int pTo){if(iFrom > iTo || pFrom > pTo){return;}// set currentcurrent = new TreeNode(postorder[pTo]);// take the last one from post order , because it is the rootvar pLast = postorder[pTo];// find its index in inordervar index = -1;for(var i = iFrom;i <= iTo; i++){if(inorder[i] == pLast){index = i;break;}}// for left sub tree , inorder : [0, index) . postorder : [pFrom, pFrom + distance(index, iFrom) - 1)Build(ref current.left, inorder, iFrom, index - 1, postorder, pFrom, pFrom + index - iFrom - 1);// for right sub tree , inorder : [index + 1, len), postorder : [pFrom + distance(index, iFrom), pTo-1]Build(ref current.right, inorder, index + 1, iTo ,postorder, pFrom + index - iFrom, pTo - 1);}}