题目链接/文章讲解/视频讲解：https://programmercarl.com/0226.%E7%BF%BB%E8%BD%AC%E4%BA%8C%E5%8F%89%E6%A0%91.html

1.这个题因为昨天的二叉树遍历的部分难度不低，然后自己昨天把那四个题都写了之后今天想巩固一下，所以这个题写了四个方法，到时候二刷的时候再来一遍。今天收获也很多。

CPP:递归

/**
 * 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 root;
        }
        swap(root->left,root->right);
        invertTree(root->left);
        invertTree(root->right);
        return root;
    }
};

CPP:前序迭代

/**
 * 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 root;
        stack<TreeNode*>st;
        st.push(root);
        while(!st.empty()){
            TreeNode *node = st.top();
            st.pop();
            swap(node->left,node->right);
            
            if(node->right)st.push(node->right);
            if(node->left)st.push(node->left);
        }
        return root;
    }
};

CPP:前序统一迭代

/**
 * 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) {
        stack<TreeNode*>st;
        if(root == NULL)return root;
        
        st.push(root);
        while(!st.empty()){
            TreeNode *node = st.top();
            if(node != NULL){
                st.pop();
                if(node->right)st.push(node->right);
                if(node->left)st.push(node->left);
                st.push(node);
                st.push(NULL);
            }else{
                st.pop();
                node = st.top();
                st.pop();
                swap(node->left,node->right);
            }
        }
        return root;
    }
};

CPP:层序，BFS

/**
 * 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) {
        queue<TreeNode*>q;
        if(root != NULL)q.push(root);
        while(!q.empty()){
            int size = q.size();
            while(size--){
                TreeNode *node = q.front();
                q.pop();
                swap(node->left,node->right);
                if(node->left)q.push(node->left);
                if(node->right)q.push(node->right);
            }
        }
        return root;
    }
};