C++生成前序遍历和中序遍历恢复二叉树

题目描述：

输入一棵二叉树的前序遍历和中序遍历，输出该二叉树的后序遍历。

解题思路：

根据二叉树的前序遍历和中序遍历可以确定一棵二叉树。前序遍历的第一个节点是根节点，根据中序遍历可以得到左子树和右子树的节点个数，从而可以将前序遍历和中序遍历分成左子树和右子树的两部分，递归构建左子树和右子树。

递归终止条件为前序遍历和中序遍历为空，此时返回空节点。

C++代码：

#include <iostream>
#include <vector>
#include <unordered_map>

using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;

    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        unordered_map<int, int> map;
        for (int i = 0; i < inorder.size(); i++) {
            map[inorder[i]] = i;
        }
        return buildTree(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1, map);
    }

private:
    TreeNode* buildTree(vector<int>& preorder, int preStart, int preEnd, vector<int>& inorder, int inStart, int inEnd, unordered_map<int, int>& map) {
        if (preStart > preEnd || inStart > inEnd) {
            return NULL;
        }
        int rootVal = preorder[preStart];
        int rootIndex = map[rootVal];
        int leftNodes = rootIndex - inStart;
        TreeNode* root = new TreeNode(rootVal);
        root->left = buildTree(preorder, preStart + 1, preStart + leftNodes, inorder, inStart, rootIndex - 1, map);
        root->right = buildTree(preorder, preStart + leftNodes + 1, preEnd, inorder, rootIndex + 1, inEnd, map);
        return root;
    }
};

void postorder(TreeNode* root) {
    if (root == NULL) {
        return;
    }
    postorder(root->left);
    postorder(root->right);
    cout << root->val << " ";
}

int main() {
    vector<int> preorder = {1, 2, 4, 5, 3, 6};
    vector<int> inorder = {4, 2, 5, 1, 3, 6};
    Solution solution;
    TreeNode* root = solution.buildTree(preorder, inorder);
    postorder(root);
    return 0;
}

输出结果为：

4 5 2 6 3 1