[LeetCode] 105. Construct Binary Tree from Preorder and Inorder Traversal 由先序和中序遍历建立二叉树

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

Note:
You may assume that duplicates do 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  20
    /  \
   15   7

给一棵树的先序和中序遍历，构建二叉树。

Java:

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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 > preorder.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

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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:

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class TreeNode:     def __init__(self, x):         self.val = x         self.left = None         self.right = None   class Solution:     # @param preorder, a list of integers     # @param inorder, a list of integers     # @return a tree node     def buildTree(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, inorder, 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)         return node   if __name__ ==  "__main__":     preorder = [1, 2, 3]     inorder = [2, 1, 3]     result = Solution().buildTree(preorder, inorder)     print(result.val)     print(result.left.val)     print(result.right.val)

C++:

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// 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 由中序和后序遍历建立二叉树

　

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