105. 从前序与中序遍历序列构造二叉树

难度中等1404收藏分享切换为英文接收动态反馈

给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。

示例 1:

输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
输出: [3,9,20,null,null,15,7]

示例 2:

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

提示:

• 1 <= preorder.length <= 3000
• inorder.length == preorder.length
• -3000 <= preorder[i], inorder[i] <= 3000
• preorder 和 inorder 均 无重复 元素
• inorder 均出现在 preorder
• preorder 保证 为二叉树的前序遍历序列
• inorder 保证 为二叉树的中序遍历序列

 

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {unordered_map<int,int>inorderId;
public:TreeNode* buildTree(vector<int>& preorder,vector<int>& inorder,int preorder_left,int preorder_right,int inorder_left,int inorder_right){if(preorder_left>preorder_right) return nullptr;int rootValue=preorder[preorder_left];int root_inorder=inorderId[rootValue];int leftTreeSize=root_inorder-inorder_left;TreeNode* newRoot=new TreeNode(rootValue);newRoot->left=buildTree(preorder,inorder,preorder_left+1,preorder_left+leftTreeSize,inorder_left,root_inorder-1);newRoot->right=buildTree(preorder,inorder,preorder_left+leftTreeSize+1,preorder_right,root_inorder+1,inorder_right);return newRoot;}TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {int n=preorder.size();for(int i=0;i<n;i++) inorderId[inorder[i]]=i;return buildTree(preorder,inorder,0,n-1,0,n-1);}
};