剑指offer学习笔记 二叉搜索树的第K大节点

面试题54：二叉搜索树的第K大节点。给定一棵二叉搜索树，找出其中第K大的节点。

中序遍历到第K个元素即可：

#include <iostream>
using namespace std;

struct BinaryTreeNode {
    int m_nValue;
    BinaryTreeNode* m_pLeft;
    BinaryTreeNode* m_pRight;
};

BinaryTreeNode* ConstructCore(int* startPreorder, int* endPreorder, int* startInorder, int* endInorder) {    //创建树
    BinaryTreeNode* root = new BinaryTreeNode();
    root->m_nValue = *startPreorder;
    root->m_pLeft = root->m_pRight = nullptr;

    if (startPreorder == endPreorder) {    //当递归到前序遍历只含一个元素时
        if (startInorder == endInorder && *startPreorder == *startInorder) {    //此时若中序遍历也只含一个元素并且这个元素值与前序遍历的元素值相同时，递归到底成功返回
            return root;
        }
        else {    //否则当前序和中序遍历的个数不相等（前序遍历只含一个元素但中序遍历有多个元素）或值不相等（前序遍历和中序遍历元素数都为1但这两个值不等）时
            throw exception("Invalid input.");    //说明输入的前序和中序遍历不匹配
        }
    }

    int* rootInorder = startInorder;
    while (rootInorder < endInorder && *rootInorder != *startPreorder) {    //遍历中序序列找到根节点
        ++rootInorder;
    }

    if (rootInorder == endInorder && *rootInorder != *startPreorder) {    //若以上循环结束仍未找到根节点，说明输入有误
        throw exception("Invalid input");
    }

    int leftLength = rootInorder - startInorder;    //左子树长度
    int* leftPreorderEnd = startPreorder + leftLength;    //左子树先序遍历尾边界
    if (leftLength > 0) {    //当左子树仍存在时
        root->m_pLeft = ConstructCore(startPreorder + 1, leftPreorderEnd, startInorder, rootInorder - 1);    //继续递归左子树
    }
    if (leftLength < endPreorder - startPreorder) {    //左子树长度小于当前遍历的树节点数-1（去掉根节点）时，说明存在右子树
        root->m_pRight = ConstructCore(startPreorder + leftLength + 1, endPreorder, rootInorder + 1, endInorder);    //继续递归右子树
    }

    return root;
}

BinaryTreeNode* Construct(int* preorder, int* inorder, int length) {    //创建树
    if (preorder == nullptr || inorder == nullptr || length <= 0) {
        return nullptr;
    }

    return ConstructCore(preorder, preorder + length - 1, inorder, inorder + length - 1);
}

BinaryTreeNode* FindKthNumCore(BinaryTreeNode* pRoot, int& k) {
    if (pRoot == nullptr) {
        return nullptr;
    }

    BinaryTreeNode* target = nullptr;

    target = FindKthNumCore(pRoot->m_pLeft, k);

    if (--k == 0) {
        target = pRoot;
    }

    if (target == nullptr) {    //如果还未找到目标
        target = FindKthNumCore(pRoot->m_pRight, k);
    }

    return target;
}

BinaryTreeNode* FindKthNum(BinaryTreeNode* pRoot, int k) {
    if (pRoot == nullptr || k == 0) {
        return nullptr;
    }

    return FindKthNumCore(pRoot, k);
}

int main() {
    int preorder[] = { 5,3,2,4,7,6,8 };
    int inorder[] = { 2,3,4,5,6,7,8 };
    BinaryTreeNode* pRoot = Construct(preorder, inorder, sizeof(preorder) / sizeof(*preorder));
    BinaryTreeNode* pRes = FindKthNum(pRoot, 4);
    if (pRes) {
        cout << pRes->m_nValue << endl;
    }
    else {
        cout << "输入错误" << endl;
    }
}