[LeetCode] 二叉树相关题目（不完全）
最近在做LeetCode上面有关二叉树的题目，这篇博客仅用来记录这些题目的代码。
二叉树的题目，一般都是利用递归来解决的，因此这一类题目对理解递归很有帮助。
1.Symmetric Tree（https://leetcode.com/problems/symmetric-tree/description/）
class Solution { public: bool isSymmetric(TreeNode* root) { return root == NULL || isMirror(root->left, root->right); } bool isMirror(TreeNode *left, TreeNode *right) { if (left == NULL && right == NULL) return true; if (left == NULL || right == NULL) return false; if (left->val != right->val) return false; return isMirror(left->left, right->right) && isMirror(left->right, right->left); } };
2.Binary Tree Level Order Traversal（https://leetcode.com/problems/binary-tree-level-order-traversal/description/）
class Solution { public: vector<vector<int>> levelOrder(TreeNode* root) { vector<vector<int>> res; if (root == NULL) return res; queue<TreeNode*> q; q.push(root); while (!q.empty()) { int layer_size = q.size(); vector<int> layer; for (int i = 0; i < layer_size; i++) { TreeNode *temp = q.front(); q.pop(); layer.push_back(temp->val); if (temp->left != NULL) q.push(temp->left); if (temp->right != NULL) q.push(temp->right)； } } return res; } };
3.Binary Tree Level Order Traversal II（https://leetcode.com/problems/binary-tree-level-order-traversal-ii/description/）
class Solution { public: vector<vector<int>> levelOrderBottom(TreeNode* root) { vector<vector<int>> res; if (root == NULL) return res; vector<int> layer; queue<TreeNode*> q; q.push(root); int count = 1; while (!q.empty()) { TreeNode* temp = q.front(); q.pop(); count--; layer.push_back(temp->val); if (temp->left != NULL) q.push(temp->left); if (temp->right != NULL) q.push(temp->right); if (count == 0) { count = q.size(); res.push_back(layer); layer.clear(); } } for (int i = 0; i < res.size() / 2; i++) { vector<int> temp = res[i]; res[i] = res[res.size() - i - 1]; res[res.size() - i - 1] = temp; } return res; } };
4.Same Tree（https://leetcode.com/problems/same-tree/description/）
class Solution { public: bool isSameTree(TreeNode* p, TreeNode* q) { if (p == NULL && q == NULL) return true; if (p == NULL || q == NULL) return false; if (p->val != q->val) return false; return isSameTree(p->left, q->left) && isSameTree(p->right, q->right); } };
5.Path Sum（https://leetcode.com/problems/path-sum/description/）
class Solution { public: bool hasPathSum(TreeNode* root, int sum) { if (root == NULL) return false; if (root->left == NULL && root->right == NULL && sum == root->val) return true; else return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val); } };
6.Path Sum II（https://leetcode.com/problems/path-sum-ii/description/）
class Solution { public: void helper(TreeNode *root, vector<vector<int>>& res, vector<int> t, int sum) { if (root == NULL) return; if (sum - root->val == 0 && root->left == NULL && root->right == NULL) { t.push_back(root->val); res.push_back(t); } else if (root->left == NULL && root->right == NULL) { return; } else { t.push_back(root->val); helper(root->left, res, t, sum - root->val); helper(root->right, res, t, sum - root->val); } } vector<vector<int>> pathSum(TreeNode* root, int sum) { vector<vector<int>> res; if (root == NULL) return res; vector<int> t; helper(root, res, t, sum); return res; } };
7.Sum Root to Leaf Numbers（https://leetcode.com/problems/sum-root-to-leaf-numbers/description/）
class Solution { public: int sumNumbers(TreeNode* root) { return helper(root, 0); } int helper(TreeNode *root, int sum) { if (root == NULL) return 0; if (root->left == NULL && root->right == NULL) return sum * 10 + root->val; else return helper(root->left, sum * 10 + root->val ) + helper(root->right, sum * 10 + root->val); } };
8.Binary Tree Right Side View（https://leetcode.com/problems/binary-tree-right-side-view/description/）
class Solution { public: vector<int> rightSideView(TreeNode* root) { vector<int> res; if (root == NULL) return res; queue<TreeNode*> q; q.push(root); while (!q.empty()) { int size = q.size(); for (int i = 0; i < size; i++) { TreeNode * temp = q.front(); q.pop(); if (temp->left) q.push(temp->left); if (temp->right) q.push(temp->right); if (i == size - 1) { res.push_back(temp->val); } } } return res; } };
9.Maximum Depth of Binary Tree（https://leetcode.com/problems/maximum-depth-of-binary-tree/description/）
递归解法：
class Solution { public: int maxDepth(TreeNode* root) { if (root == NULL) return 0; if (root->left == NULL && root->right == NULL) return 1; int maxl = maxDepth(root->left); int maxr = maxDepth(root->right); return maxl > maxr ? maxl + 1 : maxr + 1; } };
BFS解法：
class Solution { public: int maxDepth(TreeNode* root) { int res = 0; if (root == NULL) return 0; queue<TreeNode*> q; q.push(root); while (!q.empty()) { int size = q.size(); for (int i = 0; i < size; i++) { TreeNode* temp = q.front(); q.pop(); if (temp->left) q.push(temp->left); if (temp->right) q.push(temp->right); } res++; } return res; } };
10.Binary Tree Inorder Traversal(https://leetcode.com/problems/binary-tree-inorder-traversal/description/)
普通的中序遍历。
class Solution { public: vector<int> inorderTraversal(TreeNode* root) { vector<int> res; inOrder(res, root); return res; } void inOrder(vector<int>& inorder, TreeNode *root) { if (root == NULL) return; inOrder(inorder, root->left); inorder.push_back(root->val); inOrder(inorder, root->right); } };
11.Validate Binary Search（https://leetcode.com/problems/validate-binary-search-tree/）
判断一棵二叉树是否二叉搜索树。
class Solution { public: bool isValidBST(TreeNode* root) { if (root == NULL) return true; vector<int> seq; inOrder(seq, root); for (int i = 0; i < seq.size() - 1; i++) { if (seq[i] >= seq[i + 1]) return false; } return true; } void inOrder(vector<int> &seq, TreeNode *root) { if (root == NULL) return; else { inOrder(seq, root->left); seq.push_back(root->val); inOrder(seq, root->right); } } };
12.Convert Sorted Array to Binary Search Tree（https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/description/）
使用分治法。
class Solution { public: TreeNode* sortedArrayToBST(vector<int>& nums) { return BSTNode(nums, 0, nums.size() - 1); } TreeNode* BSTNode(vector<int> &nums, int first, int last) { if (first > last) return NULL; else { int mid = (first + last) / 2; TreeNode *root = new TreeNode(nums[mid]); root->left = BSTNode(nums, first, mid - 1); root->right = BSTNode(nums, mid + 1, last); return root; } } };
13.Balanced Binary Tree（https://leetcode.com/problems/balanced-binary-tree/description/）
判断一棵二叉树是否平衡二叉树。
/** * 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: bool isBalanced(TreeNode* root) { int depth = 0; return isBalanced(root, depth); } bool isBalanced(TreeNode* root, int &depth) { if (root == NULL) { depth = 0; return true; } else { int LeftDepth; int RightDepth; bool isLeftBalanced = isBalanced(root->left, LeftDepth); bool isRightBalanced = isBalanced(root->right, RightDepth); if (isLeftBalanced && isRightBalanced) { if (abs(LeftDepth - RightDepth) <= 1) { depth = LeftDepth > RightDepth ? LeftDepth + 1 : RightDepth + 1; return true; } } return false; } } };
