112. 路径总和
/*** Definition for a binary tree node.* struct TreeNode {* int val;* TreeNode *left;* TreeNode *right;* TreeNode() : val(0), left(nullptr), right(nullptr) {}* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:void preOrder(TreeNode* node, bool &flag, int Sum, int &targetSum){if (node -> left == nullptr && node -> right == nullptr){Sum += node->val;if (Sum == targetSum)flag = true;return;}if(node->left)preOrder(node->left, flag, Sum + node -> val, targetSum );if(node -> right)preOrder(node -> right, flag, Sum + node -> val, targetSum);}bool hasPathSum(TreeNode* root, int targetSum) {if(root == nullptr) return false;bool flag = false;int Sum = 0;preOrder(root, flag, Sum, targetSum);return flag;}
};
113. 路径总和 II
/*** Definition for a binary tree node.* struct TreeNode {* int val;* TreeNode *left;* TreeNode *right;* TreeNode() : val(0), left(nullptr), right(nullptr) {}* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:void preOrder(TreeNode *node, int targetSum, vector<int> &path, vector<vector<int>> &result, int pathSum){if(node -> left == nullptr && node -> right == nullptr){if(pathSum == targetSum)result.push_back(path);return;}if (node -> left){path.push_back(node -> left -> val);preOrder(node -> left, targetSum, path, result, pathSum + node -> left -> val);path.pop_back();} if (node -> right){path.push_back(node -> right -> val);preOrder(node -> right, targetSum, path, result, pathSum + node -> right -> val);path.pop_back();} }vector<vector<int>> pathSum(TreeNode* root, int targetSum) {vector<vector<int>> result;vector<int> path;if (root == nullptr) return result;path.push_back(root -> val);int pathSum = root -> val;preOrder(root, targetSum, path, result, pathSum);return result;}
};
106. 从中序与后序遍历序列构造二叉树
/*** Definition for a binary tree node.* struct TreeNode {* int val;* TreeNode *left;* TreeNode *right;* TreeNode() : val(0), left(nullptr), right(nullptr) {}* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {if (postorder.size() == 0) return nullptr;TreeNode *root = new TreeNode(postorder[postorder.size() - 1]);if (postorder.size() == 1) return root;int index = 0;for (; index < inorder.size(); index++){if(inorder[index] == root -> val)break;}vector<int> leftInorder(inorder.begin(), inorder.begin() + index);vector<int> rightInorder(inorder.begin() + index + 1, inorder.end());vector<int> leftPostorder(postorder.begin(), postorder.begin() + leftInorder.size());vector<int> rightPostorder(postorder.begin() + leftInorder.size(), postorder.end() - 1);root -> left = buildTree(leftInorder, leftPostorder);root -> right = buildTree(rightInorder, rightPostorder);return root;}
};
654. 最大二叉树
/*** Definition for a binary tree node.* struct TreeNode {* int val;* TreeNode *left;* TreeNode *right;* TreeNode() : val(0), left(nullptr), right(nullptr) {}* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:TreeNode* constructMaximumBinaryTree(vector<int>& nums) {if(nums.size() == 0) return nullptr;int index = 0;int ref = 0;for(int i = 0; i < nums.size(); i++){if(ref < nums[i]){ref = nums[i];index = i;}}TreeNode* root = new TreeNode(nums[index]);vector<int> leftnums(nums.begin(), nums.begin() + index);vector<int> rightnums(nums.begin() + index + 1, nums.end());root -> left = constructMaximumBinaryTree(leftnums);root -> right = constructMaximumBinaryTree(rightnums);return root;}
};
