翻译

将下图中上面的二叉树转换为以下的形式。详细为每一个左孩子节点和右孩子节点互换位置。

原文

如上图

分析

每次关于树的题目出错都在于边界条件上……所以这次细致多想了一遍：

void swapNode(TreeNode* tree) {
    if (tree == NULL || (tree->left == NULL && tree->right == NULL)) {}
    else    if (tree->left == NULL && tree->right != NULL) {
        TreeNode* temp = tree->right;
        tree->left = temp;
        tree->right = nullptr;
    }
    else    if (tree->right == NULL && tree->left != NULL) {
        TreeNode* temp = tree->left;
        tree->right = temp;
        tree->left = nullptr;
    }
    else {
        TreeNode* temp = tree->left;
        tree->left = tree->right;
        tree->right = temp;
    }
}            

不过这样还不够，它不过互换了一次。所以我们要用到递归：

if(tree->left != NULL)
    swapNode(tree->left);
if(tree->right != NULL)
    swapNode(tree->right);

最后在题目给定的函数内部调用自己写的这个递归函数就好。

TreeNode* invertTree(TreeNode* root) {
    if (root == NULL) return NULL;
    swapNode(root);
    return root;
}

代码

/**
* 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:
    void swapNode(TreeNode* tree) {
        if (tree == NULL || (tree->left == NULL && tree->right == NULL)) {}
        else    if (tree->left == NULL && tree->right != NULL) {
            TreeNode* temp = tree->right;
            tree->left = temp;
            tree->right = nullptr;
        }
        else    if (tree->right == NULL && tree->left != NULL) {
            TreeNode* temp = tree->left;
            tree->right = temp;
            tree->left = nullptr;
        }
        else {
            TreeNode* temp = tree->left;
            tree->left = tree->right;
            tree->right = temp;
        }
        if (tree->left != NULL)
            swapNode(tree->left);
        if (tree->right != NULL)
            swapNode(tree->right);
    }

    TreeNode* invertTree(TreeNode* root) {
        if (root == NULL) return NULL;
        swapNode(root);
        return root;
    }
};

学习

自己的解决方式还是太0基础，所以来学习学习大神的解法：

TreeNode* invertTree(TreeNode* root) {

    if(nullptr == root) return root;

    queue<TreeNode*> myQueue;   // our queue to do BFS
    myQueue.push(root);         // push very first item - root

    while(!myQueue.empty()){    // run until there are nodes in the queue 
        TreeNode *node = myQueue.front();  // get element from queue
        myQueue.pop();                     // remove element from queue

        if(node->left != nullptr){         // add left kid to the queue if it exists
            myQueue.push(node->left);
        }
        if(node->right != nullptr){        // add right kid 
            myQueue.push(node->right);
        }

        // invert left and right pointers      
        TreeNode* tmp = node->left;
        node->left = node->right;
        node->right = tmp;

    }
    return root;
} 

争取以后少用递归了。加油！