每日一题day5：[LeetCode：(二叉树专题)：剑指Offer 55 - II、07、55

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每日一题day5：LeetCode

剑指 Offer 55 - II. 平衡二叉树

剑指 Offer 07. 重建二叉树

剑指 Offer 55 - I. 二叉树的深度

剑指 Offer 28. 对称的二叉树

剑指 Offer 27. 二叉树的镜像

965. 单值二叉树

1、剑指 Offer 55 - II. 平衡二叉树

/*** 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:int GetHigh(TreeNode* root){if(root == nullptr) return 0;int l = GetHigh(root -> left);int r = GetHigh(root -> right);return max(l, r) + 1;}bool isBalanced(TreeNode* root) {if(root == nullptr) return true;int l = GetHigh(root -> left);int r = GetHigh(root -> right);return abs(l - r) <= 1 && isBalanced(root -> left) && isBalanced(root -> right);}
};

2、剑指 Offer 07. 重建二叉树

/*** 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:unordered_map<int, int> pa;TreeNode* slove(int pos, int s, int e, vector<int>& preorder, vector<int>& inorder){if(s > e) return nullptr;TreeNode* root = new TreeNode(preorder[pos]);int nxt_pos = pa[preorder[pos]];root -> left = slove(pos + 1, s, nxt_pos - 1, preorder, inorder);root -> right = slove(pos + nxt_pos - s + 1, nxt_pos + 1, e, preorder, inorder);return root;}TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {int size = preorder.size();for(int i = 0; i < size; i++) pa[inorder[i]] = i;return slove(0, 0, size - 1, preorder, inorder);}
};

3、剑指 Offer 55 - I. 二叉树的深度

/*** 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:int GetHigh(TreeNode* root){if(root == nullptr) return 0;int l = GetHigh(root -> left);int r = GetHigh(root -> right);return max(l, r) + 1;}int maxDepth(TreeNode* root) {if(root == nullptr) return 0;int l = GetHigh(root -> left);int r = GetHigh(root -> right);return max(l, r) + 1;}
};

4、剑指 Offer 28. 对称的二叉树

/*** 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:bool slove(TreeNode* l, TreeNode* r){if(l == nullptr && r == nullptr) return true;else if(l != nullptr && r == nullptr) return false;else if(l == nullptr && r != nullptr) return false;else{if(l -> val != r -> val) return false;return slove(l -> left, r -> right) && slove(l -> right, r -> left);}}bool isSymmetric(TreeNode* root) {if(root == nullptr) return true;return slove(root -> left, root -> right);}
};

5、剑指 Offer 27. 二叉树的镜像

/*** 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 slove(TreeNode* root){if(root == nullptr) return;swap(root -> left, root -> right);slove(root -> left);slove(root -> right);}TreeNode* mirrorTree(TreeNode* root) {slove(root);return root;}
};

6、965. 单值二叉树

/*** 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:bool slove(TreeNode* fa, TreeNode* son){if(fa != nullptr && son == nullptr) return true;if(fa == nullptr && son == nullptr) return true;return slove(son, son -> left) && slove(son, son -> right) && (fa -> val == son -> val);}bool isUnivalTree(TreeNode* root) {if(root == nullptr) return true;return slove(root, root -> left) && slove(root, root -> right);}
};