这里写目录标题
- 前言
- [1.——965. 单值二叉树](https://leetcode.cn/problems/univalued-binary-tree/)
- [2.——222. 完全二叉树的节点个数](https://leetcode.cn/problems/count-complete-tree-nodes/)
- [3.——144. 二叉树的前序遍历](https://leetcode.cn/problems/binary-tree-preorder-traversal/)
- [4.——94. 二叉树的中序遍历](https://leetcode.cn/problems/binary-tree-inorder-traversal/description/)
- [5.——145. 二叉树的后序遍历](https://leetcode.cn/problems/binary-tree-postorder-traversal/description/)
- [6.——226. 翻转二叉树](https://leetcode.cn/problems/invert-binary-tree/description/)
- [7.——1022. 从根到叶的二进制数之和](https://leetcode.cn/problems/sum-of-root-to-leaf-binary-numbers/)
- [8.——1379. 找出克隆二叉树中的相同节点](https://leetcode.cn/problems/find-a-corresponding-node-of-a-binary-tree-in-a-clone-of-that-tree/description/)
- [9.——1302. 层数最深叶子节点的和](https://leetcode.cn/problems/deepest-leaves-sum/description/)
- [10.——654. 最大二叉树](https://leetcode.cn/problems/maximum-binary-tree/description/)
- 结语
前言
在上期（蓝色字体可以点进去）我们一起学习了初级数据结构——二叉树，那么这期我们来运用二叉树思想进行实战，巩固知识点。
1.——965. 单值二叉树
（蓝色字体可以点进去看原题）
代码题解
/**
* 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:
bool isUnivalTree(TreeNode* root) {
if(root==NULL)return true;
if(root->left){//从根节点的左子结点递归遍历完
if(root->val!=root->left->val)return false;
if(!isUnivalTree(root->left))return false;
}
if(root->right){//再从根节点的右子节点递归遍历完
if(root->val!=root->right->val)return false;
if(!isUnivalTree(root->right))return false;
}
return true;
}
};
2.——222. 完全二叉树的节点个数
/**
* 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:
int countNodes(TreeNode* root) {
if(root==NULL)return 0;
int rc=countNodes(root->right);//左子树节点个数
int lc=countNodes(root->left);//右子树节点个数
return lc+rc+1;
}
};
3.——144. 二叉树的前序遍历
/**
* 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 {
vector<int> ret;
void preorder(TreeNode* root){
if(root){
ret.push_back(root->val);
preorder(root->left);
preorder(root->right);
}
}
public:
vector<int> preorderTraversal(TreeNode* root) {
ret.clear();
preorder(root);
return ret;
}
};
4.——94. 二叉树的中序遍历
/**
* 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 {
vector<int>ret;
void inorder(TreeNode* root){
if(root){
inorder(root->left);
ret.push_back(root->val);
inorder(root->right);
}
}
public:
vector<int> inorderTraversal(TreeNode* root) {
ret.clear();
inorder(root);
return ret;
}
};
5.——145. 二叉树的后序遍历
/**
* 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 {
vector<int>ret;
void postOrder(TreeNode* root){
if(root){
postOrder(root->left);
postOrder(root->right);
ret.push_back(root->val);
}
}
public:
vector<int> postorderTraversal(TreeNode* root) {
ret.clear();
postOrder(root);
return ret;
}
};
6.——226. 翻转二叉树
/**
* 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:
TreeNode* invertTree(TreeNode* root) {
if(root==NULL)return NULL;
TreeNode* l=invertTree(root->left);//递归反转左子树
TreeNode* r=invertTree(root->right);//递归反转右子树
root->left=r;//将root的左子树变为右子树
root->right=l;//将root的右子树变为左子树
return root;
}
};
7.——1022. 从根到叶的二进制数之和
/**
* 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 {
int sum;
void sumRoot(TreeNode* root,int pre){
if(!root->left&&!root->right){//如果是叶子结点就从该叶子结点进行求和
sum+=pre*2+root->val;
}
if(root->left){//如果该节点左子结点不为空就继续访问
sumRoot(root->left,pre*2+root->val);
}
if(root->right){//与上面同理
sumRoot(root->right,pre*2+root->val);
}
}
public:
int sumRootToLeaf(TreeNode* root) {
sum=0;
sumRoot(root,0);
return sum;
}
};
8.——1379. 找出克隆二叉树中的相同节点
/**
* 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* getTargetCopy(TreeNode* original, TreeNode* cloned, TreeNode* target){
if(original==target)return cloned;
if(original->left){
TreeNode*t=getTargetCopy(original->left,cloned->left,target);//两颗二叉树都相同递归调用
if(t)return t;
}
if(original->right){
TreeNode*t=getTargetCopy(original->right,cloned->right,target);
if(t)return t;
}
return NULL;//如果做右子树都为空，上面已经判断如果original与cloned不同所以直接返回空
}
};
9.——1302. 层数最深叶子节点的和
/**
* 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 {
int maxDepth;//记录最大深度
void getMaxDepth(TreeNode* root,int depth){
if(root==NULL)return;//递归出口
if(depth>maxDepth)maxDepth=depth;
getMaxDepth(root->left,depth+1);
getMaxDepth(root->right,depth+1);
}
int sumDepthValue(TreeNode* root,int depth){
if(root==NULL)return 0;
if(depth==maxDepth)return root->val;//如果当前深度等于最大深度就累加结果
return sumDepthValue(root->left,depth+1)+sumDepthValue(root->right,depth+1);
//左子树加上右子树最大深度的和
}
public:
int deepestLeavesSum(TreeNode* root) {
getMaxDepth(root,0);
return sumDepthValue(root,0);
}
};
10.——654. 最大二叉树
这一题不太理解的这一看我这期文章
/**
* 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 {
TreeNode* dfs(vector<int>& nums,int l,int r){//其实这整个过程就是构造大顶堆的过程
if(l>r)return NULL;//递归出口，递归到当l>r是说明这是一个空数
int maxId=l;
for(int i=l+1;i<=r;i++){//找二叉树中最大值的下标
if(nums[i]>nums[maxId])maxId=i;
}
TreeNode*root=new TreeNode();//定义一个新节点，递归构造它的左右子节点
root->left=dfs(nums,l,maxId-1);
root->right=dfs(nums,maxId+1,r);
root->val=nums[maxId];//将root的值设为最大值
return root;
}
public:
TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
return dfs(nums,0,nums.size()-1);
}
};
结语
通过这期十道题的积累相信大家对二叉树的应用与理解更加深刻，希望大家还能多多刷题，积累题库。
相信大家通过本期学习初步了解二叉树，下期作品我会更新二叉树的十几道题库，我们下期一起学习二叉树的实战应用。
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