C++/105_Construct_Binary_Tree_from_Preorder_and_Inorder_Traversal.cpp

Author: xliu45

C/105_Construct_Binary_Tree_from_Preorder_and_Inorder_Traversal.cpp

@@ -0,0 +1,60 @@
+/*
+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);
+
+}