C++/105_Construct_Binary_Tree_from_Preorder_and_Inorder_Traversal.cpp

C++/105_Construct_Binary_Tree_from_Preorder_and_Inorder_Traversal.cpp

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+// 105. Construct Binary Tree from Preorder and Inorder Traversal
+/*
+Given preorder and inorder traversal of a tree, construct the binary tree.
+
+Note: You may assume that duplicates do not exist in the tree.
+
+Tag: Tree, Array, Depth-first-search
+
+Author: Xinyu Liu
+*/
+
+#include <iostream>
+#include <vector>
+using namespace std;
+
+
+//  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) {
+        return subtree(preorder, 0, preorder.size() - 1 , inorder, 0, preorder.size() - 1);
+    }
+    TreeNode* subtree(vector<int>& preorder, int begin1, int end1, vector<int>& inorder, int begin2, int end2){
+        int i;
+        if(end1 < begin1)
+            return NULL;
+        if(end1 == begin1)
+            return new TreeNode(preorder.at(begin1));
+
+        TreeNode* root = new TreeNode(preorder.at(begin1));
+        for(i = begin2; i < end2; i++){
+            if(inorder.at(i) == root->val)
+                break;
+        }
+        int left_length = i - begin2;
+        root->left = subtree(preorder, begin1 + 1, begin1 + left_length, inorder, begin2, begin2 + left_length - 1);
+        root->right = subtree(preorder, begin1 + left_length + 1, end1, inorder, begin2 + left_length + 1, end2);
+        return root;
+    }
+};
+
+void main(){
+
+    int pre[] = {1,2,3,4};
+    vector<int> preorder(pre, pre + sizeof(pre) / sizeof(int));
+    int in[] = {2,1,4,3};
+    vector<int> inorder(in, in + sizeof(in) / sizeof(int));
+    Solution sol;
+    TreeNode* root = sol.buildTree(preorder, inorder);
+
+}