从前序与中序遍历序列构造二叉树——力扣105

题目描述


法一）递归

复杂度分析

代码如下

class Solution {
private:unordered_map<int, int> index;public:TreeNode* myBuildTree(const vector<int>& preorder, const vector<int>& inorder, int preorder_left, int preorder_right, int inorder_left, int inorder_right){if(preorder_left > preorder_right){return nullptr;}int preorder_root = preorder_left;int inorder_root = index[preorder[preorder_root]];TreeNode* root = new TreeNode(preorder[preorder_root]);int size_left_subtree = inorder_root - inorder_left;root->left = myBuildTree(preorder, inorder, preorder_left+1, preorder_left+size_left_subtree, inorder_left, inorder_root-1);root->right = myBuildTree(preorder, inorder, preorder_left+size_left_subtree+1, preorder_right, inorder_root+1, inorder_right);return root;}TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {int n = preorder.size();for(int i=0; i<n; ++i){index[inorder[i]]  = i;}return myBuildTree(preorder, inorder, 0, n-1, 0, n-1);}
};

法二）迭代




复杂度分析

代码如下

class Solution {
public:TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {if (!preorder.size()) {return nullptr;}TreeNode* root = new TreeNode(preorder[0]);stack<TreeNode*> stk;stk.push(root);int inorderIndex = 0;for (int i = 1; i < preorder.size(); ++i) {int preorderVal = preorder[i];TreeNode* node = stk.top();if (node->val != inorder[inorderIndex]) {node->left = new TreeNode(preorderVal);stk.push(node->left);}else {while (!stk.empty() && stk.top()->val == inorder[inorderIndex]) {node = stk.top();stk.pop();++inorderIndex;}node->right = new TreeNode(preorderVal);stk.push(node->right);}}return root;}
};