LeetCode 226 Invert Binary Tree（转换二叉树）

来源：互联网发布：超次元矩阵怎么玩编辑：程序博客网时间：2024/06/08 06:16

翻译

将下图中上面的二叉树转换为下面的形式，具体为每个左孩子节点和右孩子节点互换位置。

这里写图片描述

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……所以这次仔细多想了一遍：

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;    }}

仅仅是这样还不够，它只是互换了一次，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

自己的解决方案还是太初级，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！

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