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Leetcode dfs Construct Binary Tree from Inorder and Postorder Traversal
Construct Binary Tree from Inorder and Postorder Traversal
Total Accepted: 14363 Total Submissions: 54254My SubmissionsGiven inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
题意:从前序遍历和中序遍历结果重构二叉树
思路1:递归
从后序遍历结果可以得到根节点
从根节点和中序遍历结果可以得到左右子树
按上面两个规律递归就可以重构二叉树了
复杂度:时间O(n) --> 总共有n个节点要构建,空间O(log n) --> 最多需要保存前面 log n个状态,每个状态需要保存的变量是O(1)
思路2:非递归 --> todo
思路2和思路1一样,不过采用非递归的方法实现
复杂度:
vector<int> _postorder; vector<int> _inorder; void dfs(int post_start, int post_end, int in_start, int in_end, TreeNode *&root){ if(post_start >= post_end)return; root = new TreeNode(_postorder[post_end - 1]); int i; for(i = in_start; i < in_end; ++i){ if(_inorder[i] == _postorder[post_end - 1]) break; } dfs(post_start, post_start + i - in_start, in_start, in_start + i, root->left); dfs(post_start + i - in_start, post_end - 1, i + 1, in_end, root->right); } TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) { TreeNode *root = NULL; _postorder = postorder, _inorder = inorder; dfs(0, postorder.size(), 0, inorder.size(), root); return root; }
Leetcode dfs Construct Binary Tree from Inorder and Postorder Traversal
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