算法刷题之二叉树

二叉树的遍历：

前序遍历： 根左右

中序遍历：左根右

后序遍历： 左右根

 

计算二叉树有多少个结点

int count(TreeNode root){
    if(root==null) return 0;
    return 1 + count(root.left)+count(root.right);
}

226. 翻转二叉树

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翻转一棵二叉树。

示例：

输入：

     4
   /   \
  2     7
 / \   / \
1   3 6   9

输出：

     4
   /   \
  7     2
 / \   / \
9   6 3   1

备注:
这个问题是受到 Max Howell 的 原问题 启发的 ：

谷歌：我们90％的工程师使用您编写的软件(Homebrew)，但是您却无法在面试时在白板上写出翻转二叉树这道题，这太糟糕了。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        //base case
        if(root==NULL) return NULL;
        TreeNode *tmp = root->left;
        root->left = root->right;
        root->right = tmp;
        //让左右子树各自交换
        invertTree(root->left);
        invertTree(root->right);

        return root;
    }
};

 

116. 填充每个节点的下一个右侧节点指针

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给定一个完美二叉树，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *next;
}

填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。

初始状态下，所有 next 指针都被设置为 NULL。

 

示例：

输入：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

输出：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}

解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    Node(int _val, Node* _left, Node* _right, Node* _next)
        : val(_val), left(_left), right(_right), next(_next) {}
};
*/

class Solution {
public:
    Node* connect(Node* root) {
        if(root==NULL) return NULL;
        connectTwoNode(root->left,root->right);
        return root;
    }
    //定义： 输入两个节点，将其链接起来
    void connectTwoNode(Node* node1,Node* node2)
    {
        if(node1==NULL || node2==NULL) return;
        // 左右根
        node1->next = node2;
        //；链接相同父节点的两个节点
        connectTwoNode(node1->left,node1->right);
        connectTwoNode(node2->left,node2->right);
        connectTwoNode(node1->right,node2->left);
    }
};

114. 二叉树展开为链表

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给定一个二叉树，原地将它展开为一个单链表。

 

例如，给定二叉树

    1
   / \
  2   5
 / \   \
3   4   6

将其展开为：

1
 \
  2
   \
    3
     \
      4
       \
        5
         \
          6

通过次数85,514提交次数120,197

/**
 * 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:
    void flatten(TreeNode* root) {
        //base case
        if (root==nullptr) return;
        flatten(root->left);
        flatten(root->right);
        //后序遍历
        //1/ 左右子树已经被拉为一条链表
        TreeNode* left = root->left;
        TreeNode* right = root->right;

        //2.将左子树变为右子树
        root->left = nullptr;
        root->right = left;
        //3. 将原来的右子树接到当前右子树末端
        TreeNode* p = root;
        while(p->right !=nullptr)
        {
            p = p->right;
        } 
        p->right = right;
    }
};

参考链接：https://mp.weixin.qq.com/s?__biz=MzAxODQxMDM0Mw==&mid=2247487126&idx=1&sn=4de13e66397bc35970963c5a1330ce18&chksm=9bd7f09eaca0798853c41fba05ad5fa958b31054eba18b69c785ae92f4bd8e4cc7a2179d7838&token=147394321&lang=zh_CN#rd