一、题目

Given two integer arrays inorder and postorder where inorder is the inorder traversal of a binary tree and postorder is the postorder traversal of the same tree, construct and return the binary tree.

Example 1:

Input: inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
Output: [3,9,20,null,null,15,7]
Example 2:

Input: inorder = [-1], postorder = [-1]
Output: [-1]

Constraints:

1 <= inorder.length <= 3000
postorder.length == inorder.length
-3000 <= inorder[i], postorder[i] <= 3000
inorder and postorder consist of unique values.
Each value of postorder also appears in inorder.
inorder is guaranteed to be the inorder traversal of the tree.
postorder is guaranteed to be the postorder traversal of the tree.

二、题解

注意切分完中序数组后，需要用postorder.resize(postorder.size() - 1);对后序数组进行修改

/*** 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 {
public:TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {if(postorder.size() == 0) return nullptr;int rootVal = postorder[postorder.size()-1];TreeNode* root = new TreeNode(rootVal);if(postorder.size() == 1) return root;//切中序数组int index = 0;for(int i = 0;i < inorder.size();i++){if(inorder[i] == rootVal){index = i;break;}}vector<int> leInorder;vector<int> riInorder;for(int i = 0;i < index;i++){leInorder.push_back(inorder[i]);}for(int i = index + 1;i < inorder.size();i++){riInorder.push_back(inorder[i]);}postorder.resize(postorder.size() - 1);//切后序数组vector<int> lePostorder;vector<int> riPostorder;for(int i = 0;i < leInorder.size();i++){lePostorder.push_back(postorder[i]);}for(int i = leInorder.size();i < postorder.size();i++){riPostorder.push_back(postorder[i]);}root->left = buildTree(leInorder,lePostorder);root->right = buildTree(riInorder,riPostorder);return root;}
};