题目226. 翻转二叉树 - 力扣(LeetCode)
/**
* 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 swapTree(TreeNode* root)
{
if(root==nullptr)
{
return ;
}
swap(root->right,root->left);
swapTree(root->right);
swapTree(root->left);
return;
}
TreeNode* invertTree(TreeNode* root) {
swapTree(root);
return root;
}
};
题目101. 对称二叉树 - 力扣(LeetCode)
/**
* 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 compareTree(TreeNode* le,TreeNode* ri)
{
if(le==nullptr&&ri==nullptr)
{
return true;;
}
if(le==nullptr&&ri!=nullptr)
{
return false;
}
if(le!=nullptr&&ri==nullptr)
{
return false;
}
if(le->val!=ri->val)
{
return false;
}
bool bleft=compareTree(le->left,ri->right);
bool bright=compareTree(le->right,ri->left);
bool ball=bleft&&bright;
return ball;
}
bool isSymmetric(TreeNode* root) {
return compareTree(root->left,root->right);
}
};
题目104. 二叉树的最大深度 - 力扣(LeetCode)
/**
* 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 depthTree(TreeNode*root)
{
if(root==nullptr)
{
return 0;
}
int left=depthTree(root->left);
int right=depthTree(root->right);
return left>right?left+1:right+1;
}
int maxDepth(TreeNode* root) {
return depthTree(root);
}
};
题目111. 二叉树的最小深度 - 力扣(LeetCode)
/**
* 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 depthTree(TreeNode* root) {
if (root == nullptr) {
return 0;
}
int left = depthTree(root->left);
int right = depthTree(root->right);
if((left==0&&right!=0))
{
return right+1;
}
if((left!=0&&right==0))
{
return left+1;
}
return left<right?left+1:right+1;
}
int minDepth(TreeNode* root) {
return depthTree(root);
}
};
最后
判断是不是对称二叉树,这个比较绕,把一个树的左子树和右子树进行递归比较,先比较当前节点内容是否一直,一直在继续从上往下比较,在比较节点是否对称
二叉树的最小深度,注意二叉树最小深度的定义,比二叉树的最大深度加了两个判断条件
翻转二叉树,就是先翻转再递归
摆烂了一周,要继续练算法了,加油