已知前序中序遍历，求二叉树的高度

 

此题所涉及的建树详细请见我上一篇博文，二叉树的构造，这种构造方法简单容易理解 

此处求树高采用的是递归求解，对每一层最长树加上当前节点的高度 

#include 
#include 
#include 
using namespace std;

struct TreeNode {
    char val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(char val) : val(val), left(nullptr), right(nullptr) {}
};

// 根据先序遍历和中序遍历序列构造二叉树
TreeNode* buildTree(string& preorder, string& inorder,
                    int preStart, int inStart, int inEnd,
                    unordered_map& indexMap) {
    if (preStart > preorder.size() - 1 || inStart > inEnd) {
        return nullptr;
    }

    char rootValue = preorder[preStart];
    TreeNode* root = new TreeNode(rootValue);
    int rootIndex = indexMap[rootValue];

    root->left = buildTree(preorder, inorder, preStart + 1, inStart, rootIndex - 1, indexMap);
    root->right = buildTree(preorder, inorder, preStart + rootIndex - inStart + 1, rootIndex + 1, inEnd, indexMap);

    return root;
}

// 计算二叉树的高度
int getHeight(TreeNode* root) {
    if (root == nullptr) {
        return 0;
    }

    int leftHeight = getHeight(root->left);
    int rightHeight = getHeight(root->right);

    return max(leftHeight, rightHeight) + 1;
}

int main() {
    int n;
    while (cin >> n) {
        string preorder, inorder;
        cin >> preorder >> inorder;

        unordered_map indexMap;
        for (int i = 0; i < n; ++i) {
            indexMap[inorder[i]] = i;
        }

        TreeNode* root = buildTree(preorder, inorder, 0, 0, n - 1, indexMap);
        int height = getHeight(root);

        cout << height << endl;
    }
    return 0;
}