2021.2.3 刷题(二叉树翻转)

题目链接：https://leetcode-cn.com/problems/invert-binary-tree/
题目描述：

方法一：层序遍历(BFS)
采用层序遍历的方式，每遍历一个节点，就将该节点的左右孩子交换。

/**
 * 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 *> que;
        if(root != nullptr)
            que.push(root);
        while(!que.empty())
        {
            TreeNode *node = que.front();
            que.pop();
            swap(node->left, node->right);  //交换左右孩子
            if(node->left) que.push(node->left);
            if(node->right) que.push(node->right);
        }
        return root;
    }
};

2.迭代法(前序遍历)DFS

/**
 * 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 != nullptr)
            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;
    }
};

3.递归法

class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
       
       if(root == nullptr)
            return root;
        swap(root->left, root->right);
        invertTree(root->left);
        invertTree(root->right);
        return root;
    }
};