The idea is very simple, just use the idea of divide and conquer. The core is to find out the array where the left and right subtrees are located.
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode* reConstructBinaryTree(vector<int> pre, int preStart, int preEnd, vector<int> vin, int vinStart, int vinEnd) { if (preStart> preEnd || vinStart> vinEnd) { return NULL; } int value = pre[preStart]; TreeNode *root = new TreeNode(value); for (int i = vinStart; i <= vinEnd; ++i) { if (value == vin[i]) { int leftTreeSize = i - vinStart; root->left = reConstructBinaryTree(pre, preStart+1, preStart + leftTreeSize, vin, vinStart, i - 1); root->right = reConstructBinaryTree(pre, preStart + leftTreeSize + 1, preEnd, vin, i + 1, vinEnd); } } return root; } TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) { int preSize = pre.size(); int vinSize = vin.size(); if (preSize == 0 || preSize! = vinSize) { return NULL; } return reConstructBinaryTree(pre, 0, preSize - 1, vin, 0, vinSize - 1); } };