标签：pst and ini 构造 代码 length 结果 class ++
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
inorder = [9,3,15,20,7] postorder = [9,15,7,20,3]Return the following binary tree:
3 / 9 20 / 15 7
从中序和后序遍历序列构造二叉树。题目就是题意，影子题105，几乎一模一样。
思路还是递归/分治。因为有了后序遍历所以后序遍历的最后一个node就是根节点。有了根节点，可以依据这个根节点在inorder遍历的位置，将inorder的结果分成左子树和右子树。照着例子解释一下，
inorder = [9,3,15,20,7] - 9是左子树，15，20，7是右子树
postorder = [9,15,7,20,3] - 3是根节点
思路和代码可以参照105题，几乎是一样的。
时间O(n)
空间O(n)
Java实现
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 class Solution { 11 public TreeNode buildTree(int[] inorder, int[] postorder) { 12 // corner case 13 if (inorder == null || postorder == null || inorder.length == 0 || postorder.length == 0) { 14 return null; 15 } 16 return helper(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1); 17 } 18 19 private TreeNode helper(int[] inorder, int inStart, int inEnd, int[] postorder, int pStart, int pEnd) { 20 // corner case 21 if (inStart > inEnd || pStart > pEnd) { 22 return null; 23 } 24 TreeNode root = new TreeNode(postorder[pEnd]); 25 int index = 0; 26 while (inorder[inStart + index] != postorder[pEnd]) { 27 index++; 28 } 29 root.left = helper(inorder, inStart, inStart + index - 1, postorder, pStart, pStart + index - 1); 30 root.right = helper(inorder, inStart + index + 1, inEnd, postorder, pStart + index, pEnd - 1); 31 return root; 32 } 33 }
相关题目
105. Construct Binary Tree from Preorder and Inorder Traversal
106. Construct Binary Tree from Inorder and Postorder Traversal
[LeetCode] 106. Construct Binary Tree from Inorder and Postorder Traversal
标签：pst and ini 构造 代码 length 结果 class ++
原文地址：https://www.cnblogs.com/cnoodle/p/12936155.html
