leetcode 106. 从中序与后序遍历序列构造二叉树

leetcode 106. 从中序与后序遍历序列构造二叉树

题目

给定两个整数数组 inorder 和 postorder ，其中 inorder 是二叉树的中序遍历， postorder 是同一棵树的后序遍历，请你构造并返回这颗 二叉树 。

示例 1:

输入：inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
输出：[3,9,20,null,null,15,7]
示例 2:

输入：inorder = [-1], postorder = [-1]
输出：[-1]

提示:

1 <= inorder.length <= 3000
postorder.length == inorder.length
-3000 <= inorder[i], postorder[i] <= 3000
inorder 和 postorder 都由 不同 的值组成
postorder 中每一个值都在 inorder 中
inorder 保证是树的中序遍历
postorder 保证是树的后序遍历

思路

其实之前面试折过一次，已经大致记得思路了，不过实现起来其实是有些坑的，就是不断递归，一开始打算递归到只有一个节点时候结束，返回一个TreeNode。具体切分区间的方式利用了两个map，通过后序遍历的最后一个作为根节点找中序遍历切分点idx，然后根据中序遍历切分的两块区间去找后序遍历下一节点的两个区间。
但这种实现方式不好判定边界值。一旦idx是头或者尾就会报错，所以我们采用的寻找区间方式是利用idx与中序遍历区间的头节点差值作为左侧还有多少节点待插入。然后每次找到根节点再创建root，这样避免了很多麻烦

题解

// 错误的
/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {Map<Integer, Integer> inMap = new HashMap<>();Map<Integer, Integer> postMap = new HashMap<>();public TreeNode buildTree(int[] inorder, int[] postorder) {for (int i=0;i<inorder.length;i++) {inMap.put(inorder[i], i);}for (int i=0;i<postorder.length;i++) {postMap.put(postorder[i], i);}TreeNode root = new TreeNode(postorder[postorder.length - 1]);handler(root, inorder, postorder, 0, inorder.length - 1, 0, postorder.length - 1);return root;}public TreeNode handler(TreeNode root, int[] inorder, int[] postorder, int inSt, int inEd, int postSt, int postEd) {if (inSt == inEd) {return new TreeNode(inorder[inSt]);}int idx = inMap.get(postorder[postEd]);System.out.println(inSt + " " + inEd + " " + postSt + " " + postEd + " " + idx);int postleftEd = postMap.get(inorder[idx - 1]);int postrightSt = postMap.get(inorder[idx + 1]);root.left = handler(new TreeNode(postorder[postleftEd]), inorder, postorder, inSt, idx - 1, postSt, postleftEd);root.right = handler(new TreeNode(postorder[postEd - 1]), inorder, postorder, idx + 1, inEd, postrightSt, postEd - 1);return root;}
}
// 正确的
/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {Map<Integer, Integer> inMap = new HashMap<>();Map<Integer, Integer> postMap = new HashMap<>();public TreeNode buildTree(int[] inorder, int[] postorder) {for (int i=0;i<inorder.length;i++) {inMap.put(inorder[i], i);}for (int i=0;i<postorder.length;i++) {postMap.put(postorder[i], i);}TreeNode root = handler(inorder, postorder, 0, inorder.length - 1, 0, postorder.length - 1);return root;}public TreeNode handler(int[] inorder, int[] postorder, int inSt, int inEd, int postSt, int postEd) {if (inSt > inEd || postSt > postEd) {return null;}int idx = inMap.get(postorder[postEd]);TreeNode root = new TreeNode(postorder[postEd]);int leftNum = idx - inSt;root.left = handler(inorder, postorder, inSt, idx - 1, postSt, postSt + leftNum - 1);root.right = handler(inorder, postorder, idx + 1, inEd, postSt + leftNum, postEd - 1);return root;}
}