LeetCode226 翻转二叉树

原题目

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

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/invert-binary-tree

题目分析

只要将每个节点的左右子树交换即可
前序，中序，后序，层序都可以

完整代码

前序递归

struct TreeNode* invertTree(struct TreeNode* root){
    if(root==NULL)return root;
    struct TreeNode *temp=root->right;
    root->right=root->left;
    root->left=temp;
    invertTree(root->left);
    invertTree(root->right);
    return root;
}

前序迭代

struct TreeNode* invertTree(struct TreeNode* root){
    if(root==NULL)return root;
    struct TreeNode* stack[200];
    int top=0;
    stack[top++]=root;
    struct TreeNode *cur;
    while(top)
    {
        cur=stack[--top];
        if(cur->right)
            stack[top++]=cur->right;
        if(cur->left)
            stack[top++]=cur->left;
        struct TreeNode *temp=cur->left;
        cur->left=cur->right;
        cur->right=temp;
    }
    return root;
}

中序递归

struct TreeNode* invertTree(struct TreeNode* root){
    if(root==NULL)return root;
    invertTree(root->left);
    struct TreeNode *temp=root->right;
    root->right=root->left;
    root->left=temp;
    invertTree(root->left);
    
    return root;
}

层序迭代

struct TreeNode* invertTree(struct TreeNode* root){
    if(root==NULL)return root;
    struct TreeNode* queue[200];
    int front=0,rear=0;
    queue[rear++]=root;
    struct TreeNode *cur;
    while(front!=rear)
    {
        cur=queue[front++];
        if(cur->left)
            queue[rear++]=cur->left;
        if(cur->right)
            q ueue[rear++]=cur->right;
        struct TreeNode *temp=cur->left;
        cur->left=cur->right;
        cur->right=temp;
    }
    return root;
}