[Leetcode] 105. Construct binary tree from preorder and inorder traversal by first order and mid-order traversal build two forks

Given Preorder and inorder traversal of a tree, construct the binary tree.

Note:
Assume that duplicates does not exist in the tree.

For example, given

preorder = [3,9,20,15,7]inorder = [9,3,15,20,7]

Return the following binary tree:

    3   /   9    /   7

A binary tree is constructed by traversing the first and middle sequence of a tree.

Java:

Public TreeNode Buildtree (int[] preorder, int[] inorder) {    return helper (0, 0, inorder.length-1, preorder, inorder); }public TreeNode Helper (int prestart, int instart, int inend, int[] preorder, int[] inorder) {    if (Prestart > Preor der.length-1 | | Instart > Inend) {        return null;    }    TreeNode root = new TreeNode (Preorder[prestart]);    int inindex = 0; Index of current root in inorder for    (int i = Instart; I <= inend; i++) {        if (inorder[i] = = root.val) {            Inindex = i;        }    }    Root.left = Helper (Prestart + 1, Instart, inIndex-1, preorder, inorder);    Root.right = Helper (Prestart + Inindex-instart + 1, Inindex + 1, inend, preorder, inorder);    return root;}

Python:recursive

def buildtree (self, Preorder, inorder):    if inorder:        ind = inorder.index (Preorder.pop (0))        root = TreeNode ( Inorder[ind])        root.left = Self.buildtree (preorder, Inorder[0:ind])        root.right = Self.buildtree (Preorder, Inorder[ind+1:])        return root

Python:

Class Treenode:def __init__ (self, x): Self.val = x self.left = None self.right = Noneclass Soluti On: # @param preorder, a list of integers # @param inorder, a list of integers # @return a tree node def build        Tree (self, Preorder, inorder): Lookup = {} for I, Num in enumerate (inorder): lookup[num] = i Return Self.buildtreerecu (lookup, preorder, inorder, 0, 0, Len (inorder)) def buildtreerecu (self, lookup, preorder, I Norder, Pre_start, In_start, in_end): if In_start = = In_end:return None node = TreeNode (Preorder        [Pre_start]) i = Lookup[preorder[pre_start]] node.left = SELF.BUILDTREERECU (lookup, preorder, inorder, Pre_start + 1, In_start, i) Node.right = SELF.BUILDTREERECU (lookup, preorder, inorder, Pre_start + 1 + i-in_start, i + 1, in_end) r Eturn Nodeif __name__ = = "__main__": preorder = [1, 2, 3] inorder = [2, 1, 3] result = solution (). Buildtree (Preo Rder, inorder) print (result.val) print (result.left.val) print (result.right.val) 

C + +:

Time:o (n)//Space:o (n)/** * 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:treenode* Buildtree (vector<int>& preorder, vector<int>& inorder) {        Unordered_map<int, size_t> In_entry_idx_map;        for (size_t i = 0; i < inorder.size (); ++i) {In_entry_idx_map.emplace (inorder[i], i);                                            } return Reconstructpreinordershelper (preorder, 0, Preorder.size (), inorder, 0, Inorder.size (),    IN_ENTRY_IDX_MAP);    }//reconstructs the binary tree from pre[pre_s:pre_e-1] and//in[in_s:in_e-1].                                           TreeNode *reconstructpreinordershelper (const vector<int>& Preorder, size_t pre_s, size_t pre_e,                    Const vector<int>& Inorder, size_t in_s, size_t in_e,                       Const Unordered_map<int, size_t>& in_entry_idx_map) {if (pre_s = = Pre_e | | in_s =        = in_e) {return nullptr;        } Auto idx = in_entry_idx_map.at (preorder[pre_s]);        Auto left_tree_size = idx-in_s;        Auto node = new TreeNode (preorder[pre_s]);                                                  Node->left = Reconstructpreinordershelper (preorder, pre_s + 1, pre_s + 1 + left_tree_size,        Inorder, in_s, IDX, in_entry_idx_map);                                                   Node->right = Reconstructpreinordershelper (preorder, pre_s + 1 + left_tree_size, pre_e,        Inorder, idx + 1, in_e, in_entry_idx_map);    return node; }};

　　

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[Leetcode] 106. Construct binary tree from inorder and postorder traversal by the middle sequence and post-order traversal to build two-fork trees

　

[Leetcode] 105. Construct binary tree from preorder and inorder traversal by first order and mid-order traversal build two forks