我要好offer之 二叉树大总结

一. 二叉树定义

二叉树具有天然的递归特性,凡是二叉树相关题,首先应该联想到递归

struct BinTreeNode {
    BinTreeNode* left;
    BinTreeNode* right;
    int          val;
    BinTreeNode(int value) : left(nullptr), right(nullptr), val(value) { }
};

 

二. 二叉树遍历

详见笔者博文:二叉树遍历大总结

#include <iostream>
#include <vector>
#include <stack>
using namespace std;

// 二叉树递归定义
struct BinTreeNode {
    BinTreeNode* left;
    BinTreeNode* right;
    int          val;
    BinTreeNode(int value) : left(NULL), right(NULL), val(value) { }
};

// 根据结点值构建二叉树
BinTreeNode* BuildBinTree(BinTreeNode* root, int value) {
    if (root == NULL) {
        root = new BinTreeNode(value);  // 结点为空,new一个节点
    } else if (value <= root->val) {
        root->left = BuildBinTree(root->left, value);
    } else {
        root->right = BuildBinTree(root->right, value);
    }
    return root;
}

// 先序遍历,递归
void PreOrder(BinTreeNode* root) {
    if (root == NULL) return;
    std::cout << root->val << "-->";  // 访问节点
    PreOrder(root->left);
    PreOrder(root->right);
}

// 先序遍历,非递归,栈版本
void PreOrderStk(BinTreeNode* root) {
    if (root == NULL) return;
    std::stack<BinTreeNode*> stk;
    while (root != NULL || !stk.empty()) {
        while (root != NULL) {
            std::cout << root->val << "-->";  // 访问节点
            stk.push(root);
            root = root->left;  // 深度优先搜索到最左节点,然后回溯
        }
        BinTreeNode* cur = stk.top();
        stk.pop();
        root = cur->right;     // 转向右子树
    }
}

// 先序遍历,非递归,非栈,Morris
void PreOrderMorris(BinTreeNode* root) {
    if (root == NULL) return;
    BinTreeNode* prev = NULL;  
    BinTreeNode* cur = root;   // cur为正在访问的结点
    while (cur != NULL) {
        if (cur->left == NULL) {
            std::cout << cur->val << "-->";  // 访问节点
            cur = cur->right;
        } else {
            prev = cur->left;            // 前驱结点为 左子树最右下节点
            while (prev->right != NULL && prev->right != cur) {
                prev = prev->right;
            }
            if (prev->right == NULL) {           // 还没线索化,则建立线索
                std::cout << cur->val << "-->";  // 仅这一行与中序不同
                prev->right = cur;               // 建立线索
                cur = cur->left;                 // 转向处理左子树
            } else {
                cur = cur->right;                // 转向处理右子树
                prev->right = NULL;              // 已经线索化,则删除线索
            }
        }
    }
}

// 中序遍历,递归
void InOrder(BinTreeNode* root) {
    if (root == NULL) return;
    InOrder(root->left);
    std::cout << root->val << "-->";  // 访问节点
    InOrder(root->right);
}

// 中序遍历,非递归,栈版本
void InOrderStk(BinTreeNode* root) {
    if (root == NULL) return;
    std::stack<BinTreeNode*> stk;
    while (root != NULL || !stk.empty()) {
        while (root != NULL) {
            stk.push(root);
            root = root->left;  // 深度优先搜索到最左节点,然后回溯
        }
        BinTreeNode* cur = stk.top();
        stk.pop();
        std::cout << cur->val << "-->";  // 访问节点
        root = cur->right;     // 转向右子树
    }
}

// 中序遍历,非递归,非栈,Morris
void InOrderMorris(BinTreeNode* root) {
    if (root == NULL) return;
    BinTreeNode* prev = NULL;  
    BinTreeNode* cur = root;   // cur为正在访问的结点
    while (cur != NULL) {
        if (cur->left == NULL) {
            std::cout << cur->val << "-->";  // 访问节点
            cur = cur->right;
        } else {
            prev = cur->left;               // 前驱结点为 左子树最右下节点
            while (prev->right != NULL && prev->right != cur) {
                prev = prev->right;
            }
            if (prev->right == NULL) {          // 还没线索化,则建立线索
                prev->right = cur;              // 建立线索
                cur = cur->left;                // 转向处理左子树
            } else {
                std::cout << cur->val << "-->"; // 已经线索化,则访问节点,并删除线索
                cur = cur->right;               // 转向处理右子树
                prev->right = NULL;             // 删除线索
            }
        }
    }
}

// 后序遍历,递归
void PostOrder(BinTreeNode* root) {
    if (root == NULL) return;
    PostOrder(root->left);
    PostOrder(root->right);
    std::cout << root->val << "-->";
}

// 后序遍历,非递归,栈版本
void PostOrderStk(BinTreeNode* root) {
    if (root == NULL) return;
    std::stack<BinTreeNode*> stk;
    BinTreeNode* cur = root;
    BinTreeNode* prev = NULL;
    do {
        while (cur != NULL) {
            stk.push(cur);
            cur = cur->left;
        }
        prev = NULL;
        while (!stk.empty()) {
            cur = stk.top();
            stk.pop();
            if (cur->right == prev) {   // 从右子树返回,表示其右子树已经访问完毕
                std::cout << cur->val << "-->";
                prev = cur;
            } else {
                stk.push(cur);
                cur = cur->right;      // 其右子树还未访问,转向其右子树访问
                break;
            }
        }
    } while (!stk.empty());
}

// 层次遍历,递归版
void LevelOrder(BinTreeNode* root, int level, std::vector<std::vector<int>>& res) {
    if (root == NULL) return;
    if (level > res.size()) {
        res.push_back(std::vector<int>());
    }
    res.at(level - 1).push_back(root->val);
    LevelOrder(root->left, level + 1, res);
    LevelOrder(root->right, level + 1, res);
}
std::vector<std::vector<int>> LevelOrderHelp(BinTreeNode* root) {
    std::vector<std::vector<int>> res;
    LevelOrder(root, 1, res);
    return res;
}

// 层次遍历,非递归,队列版
void LevelOrderQueue(BinTreeNode* root) {
    if (root == NULL) return;
    std::deque<BinTreeNode*> dequeTreeNode;
    dequeTreeNode.push_back(root);
    while (!dequeTreeNode.empty()) {
        BinTreeNode* cur = dequeTreeNode.front();
        dequeTreeNode.pop_front();
        std::cout << cur->val << "-->"; // 访问节点
        if (cur->left  != NULL) dequeTreeNode.push_back(cur->left);
        if (cur->right != NULL) dequeTreeNode.push_back(cur->right);
    }
}

int main() {
    // your code goes here
    std::vector<int> num = {6, 2, 7, 1, 4, 9, 3, 5, 8};
    BinTreeNode* root = nullptr;
    for (auto val : num) {
        root = BuildBinTree(root, val);
    }
    std::cout << "1:PreOrder:" << std::endl;
    std::cout << "\t PreOrder:" ; 
    PreOrder(root);
    std::cout << std::endl;
    std::cout << "\t PreOrderStk:"; 
    PreOrderStk(root); 
    std::cout << std::endl;
    std::cout << "\t PreOrderMorris:"; 
    PreOrderMorris(root); 
    std::cout << std::endl;
    
    std::cout << "2:InOrder:" << std::endl;
    std::cout << "\t InOrder:";
    InOrder(root); 
    std::cout << std::endl;
    std::cout << "\t InOrderStk:";
    InOrderStk(root); 
    std::cout << std::endl;
    std::cout << "\t InOrderMorris:";
    InOrderMorris(root);
    std::cout << std::endl;
    
    std::cout << "3:PostOrder:" << std::endl;
    std::cout << "\t PostOrder:";
    PostOrder(root);
    std::cout << std::endl;
    std::cout << "\t PostOrderStk:";
    PostOrderStk(root);
    std::cout << std::endl;

    std::cout << "4:LevelOrder:" << std::endl;
    std::cout << "\t LevelOrder:";
    LevelOrderQueue(root); 
    std::cout << std::endl;
    return 0;
}

 

三. 二叉树构建

1. 先序序列和中序序列构建一颗二叉树  中序序列和后序序列构建一棵二叉树

注:为什么先序和后序不能确定一棵二叉树?(例如:先序ABCD,后序CBDA)

给定先序遍历序列和后序遍历序列,求有多少种不同形态的二叉树?

考虑简单的两个节点A和B,前序遍历为AB,后序遍历为BA,则会存在两种二叉树: A->left=B 和 A->right=B

存在不同二叉树的原因在于某节点缺少其中一个子树,在遍历序列中呈现出前序和后序遍历中节点的值相邻但顺序相反(例如先序AB,后序BA),那么总的二叉树数量番倍

int BinTreeNum(const std::vector<int>& preOrder, const std::vector<int>& postOrder) {
    assert(preOrder.size() == postOrder.size());
    int num = 1;
    for (int i = 0; i < preOrder.size(); ++i) {
        auto iter = std::find(postOrder.begin(), postOrder.end(), preOrder.at(i));
        int index = std::distance(postOrder.begin(), iter);
        if (i < preOrder.size() - 1 && index > 0 && preOrder.at(i+1) == postOrder.at(index-1)) {
            num *= 2;
        }
    }
    return num;
}

 

2. 有序数组/链表 构建一颗二叉查找树BST

 

四. 二叉树判断

 1. 判断二叉查找树BST

 2. 判断平衡二叉树

 3. 判断对称二叉树

 4. 判断相同二叉树

 5. 判断二叉树的子树

 

五. 二叉树路径相关问题

 1. 二叉树路径和问题

 

 2. 二叉树节点之间的最大距离(任意两个节点之间的最大步数)

二叉树中相距最远的两个节点之间的距离。
递归解法:
(1)如果二叉树为空,返回0,同时记录左子树和右子树的深度,都为0
(2)如果二叉树不为空,最大距离要么是左子树中的最大距离,要么是右子树中的最大距离,要么是左子树节点中到根节点的最大距离+右子树节点中到根节点的最大距离,同时记录左子树和右子树节点中到根节点的最大距离

int GetMaxDistance(BinTreeNode* root, int& maxLeft, int& maxRight) {
    // maxLeft,  左子树中的节点距离根节点的最远距离  
    // maxRight, 右子树中的节点距离根节点的最远距离
    if (root == NULL) {
        maxLeft = 0;
        maxRight = 0;
        return 0;
    }
    int maxLL, maxLR, maxRL, maxRR;
    int maxDistLeft = 0;
    int maxDistRight = 0;
    
    if (root->left != NULL) {
        maxDistLeft = GetMaxDistance(root->left, maxLL, maxLR);
        maxLeft = std::max(maxLL, maxLR) + 1;
    } else {
        maxDistLeft = 0;
        maxLeft = 0;
    }
    
    if (root->right != NULL) {
        maxDistRight = GetMaxDistance(root->right, maxRL, maxRR);
        maxRight = std::max(maxRL, maxRR) + 1;
    } else {
        maxDistRight = 0;
        maxRight = 0;
    }
    
    return std::max(maxLeft + maxRight, std::max(maxDistLeft, maxDistRight));
}

 

 3. 二叉树最大路径和问题(任意两个节点之间)

思路:有点类似于 数组最大子段和问题,部分和大于0,表示贡献值为0,可以相加起来

我们对二叉树进行 dfs,先算出 左右子树的和值 leftSum 和 rightSum, 如果 leftSum 或者 rightSum 大于0,则表示此结果对 后续结果是有利的

int maxSum = INT_MIN;
int dfs(BinTreeNode* root) {
    if (root == NULL) return 0;
    int leftSum = dfs(root->left);
    int rightSum = dfs(root->right);
    int sum = root->val;
    if (leftSum > 0)  sum += leftSum;
    if (rightSum > 0) sum += rightSum;
    maxSum = std::max(maxSum, sum);
    if (std::max(leftSum, rightSum) > 0) {
        return std::max(leftSum, rightSum) + root->val;
    } else {
        return root->val;
    }
}

int maxPathSum(BinTreeNode* root) {
    if (root == NULL) return 0;
    dfs(root);
    return maxSum;
}

 

 4. 二叉树最低公共祖先

思路:先求取 根节点到两个结点的路径序列,然后在这两个路径中查找最后一个公共结点即可

扩展:如何求 两个结点之间的距离呢? 可以先分别求得 根节点到这两个结点的路径,然后 最低公共祖先结点到 结点A的距离 + 最低公共祖先结点到 结点B的距离

struct BinTreeNode {
    BinTreeNode* left;
    BinTreeNode* right;
    int          val;
    BinTreeNode(int value) : left(NULL), right(NULL), val(value) { }
};


BinTreeNode* BuildBinTree(BinTreeNode* root, int value) {
    if (root == NULL) {
        root = new BinTreeNode(value);
    } else if (value <= root->val) {
        root->left = BuildBinTree(root->left, value);
    } else {
        root->right = BuildBinTree(root->right, value);
    }
    return root;
}

// 二叉树中序 递归
void Inorder(BinTreeNode* root) {
    if (root == NULL) return;
    Inorder(root->left);
    std::cout << root->val << "-->";
    Inorder(root->right);
}

// 二叉树中序 非递归 栈
void InorderStk(BinTreeNode* root) {
    if (root == NULL) return;
    std::stack<BinTreeNode*> stk;
    while (root != NULL || !stk.empty()) {
        while (root != NULL) {
            stk.push(root);
            root = root->left;
        }
        BinTreeNode* cur = stk.top();
        stk.pop();
        std::cout << cur->val << "-->";
        root = cur->right;
    }
}

// 二叉树中序 非递归 非栈
void InorderMorris(BinTreeNode* root) {
    if (root == NULL) return;
    BinTreeNode* cur = root;
    BinTreeNode* prev = NULL;
    while (cur != NULL) {
        if (cur->left == NULL) {
            std::cout << cur->val << "-->";
            prev = cur;
            cur = cur->right;
        } else {
            BinTreeNode* node = cur->left;
            while (node->right != NULL && node->right != cur) {
                node = node->right;
            }
            if (node->right == NULL) {  
                node->right = cur;
                cur = cur->left;
            } else {
                std::cout << cur->val << "-->";
                prev = cur;
                cur = cur->right;
                node->right = NULL;
            }
        }
    }
}

// 按值查找结点
BinTreeNode* GetNode(BinTreeNode* root, int value) {
    if (root == NULL) return NULL;
    if (root->val == value) {
        return root;
    } else if (value < root->val) {
        return GetNode(root->left, value);
    } else {
        return GetNode(root->right, value);
    }
}

// 获取从根节点到某节点的路径
bool GetNodePath(BinTreeNode* root, BinTreeNode* pNode, std::vector<BinTreeNode*>& path) {
    if (root == NULL || pNode == NULL) return false;
    bool found = false;
    if (root == pNode) {
        path.push_back(root);
        found = true;
        return found;
    }
    path.push_back(root);
    found = GetNodePath(root->left, pNode, path);
    if (!found) {  // 左子树未找到,继续右子树找
        found = GetNodePath(root->right, pNode, path);
    }
    if (!found) { // 左右子树均未找到,此时才pop
        path.pop_back();
    }
    return found;
}

// 两个结点的最低公共祖先
BinTreeNode* GetLastCommonParent(BinTreeNode* root, BinTreeNode* pNode1, BinTreeNode* pNode2) {
    if (root == NULL || pNode1 == NULL || pNode2 == NULL) {
        return NULL;
    }
    std::vector<BinTreeNode*> path1;  
    std::vector<BinTreeNode*> path2;
    bool bFound1 = GetNodePath(root, pNode1, path1);
    bool bFound2 = GetNodePath(root, pNode2, path2);
    if (bFound1 == false || bFound2 == false) {
        return NULL;
    }
    std::vector<BinTreeNode*>::iterator iter1 = path1.begin();
    std::vector<BinTreeNode*>::iterator iter2 = path2.begin();
    for (; iter1 != path2.end() && iter2 != path2.end(); ++iter1, ++iter2) {
        if (*iter1 != *iter2) {
            break;
        }
    }
    --iter1;
    return *iter1;
}

 

posted @ 2014-08-12 20:56 skyline09 阅读(...) 评论(...) 编辑 收藏