leetcode-tree1

来源：互联网发布：unity3d网络游戏开发编辑：程序博客网时间：2024/05/18 02:25

1.Invert Binary Tree 二叉树的镜像

2.Minimum Depth of Binary Tree

3.Maximum Depth of Binary Tree

struct TreeNode {    int val;    struct TreeNode *left;    struct TreeNode *right;};struct TreeNode* invertTree(struct TreeNode* root) {struct TreeNode* temp;if (root == NULL || (root->left == NULL && root->right == NULL)){return root;}//else if (root->left != NULL && root->right == NULL)//{////root->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));//root->right = root->left;//root->right->left = root->right->right = root->left = NULL;//}//else if (root->right != NULL && root->left == NULL)//{//root->left = (struct TreeNode*)malloc(sizeof(struct TreeNode));//root->left->val = root->right->val;//root->left->left = root->left->right = root->right = NULL;//}else{temp = root->left;root->left = root->right;root->right = temp;}root->left = invertTree(root->left);root->right = invertTree(root->right);}int minDepth(struct TreeNode* root) {if (root == NULL ){return 0;}else if ((root->left == NULL && root->right == NULL)){return 1;}else if (root->left == NULL){return minDepth(root->right) + 1;}else if (root->right == NULL){return minDepth(root->left) + 1;}else{return (minDepth(root->left)> minDepth(root->right) ? minDepth(root->right) : minDepth(root->left)) + 1;}}int maxDepth(struct TreeNode* root) {int left, right;if (root == NULL){return 0;}left = maxDepth(root->left);right = maxDepth(root->right);if (left == 0 && right == 0){return 1;}return (left > right ? left : right) + 1;}

这三道题都是比较简单的递归实现，但是编写时也出现了错误：

①树的镜像，左右树的交换，我最初想成了左右数值的交换，题目看懂了，但是想错了；

②第二三题可以用深度优先搜索算法（DFS），了解了一下分层遍历，需要借助队列来实现，而递归都是默认的使用栈，C语言不好实现非递归方法；

③先做了第二题才做第三题，开始只是简单的改了min为max等，提交时发现超时；递归会占用栈，数据太多时会越界、效率也比较低，可以避免尽量避免，可能的用变量代替；

④ leetcode 根的深度为1

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