Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) { int *pre = &preorder[0]; int *in = &inorder[0]; return build(pre, in, inorder.size()); } TreeNode *build(int *pre, int *in, int n) { if(n == 0) return nullptr; TreeNode *root = new TreeNode(pre[0]); int i = 0; for(; i<n; i++) { if(in[i] == pre[0]) break; } root->left = build(pre+1, in, i); root->right = build(pre+i+1, in+i+1, n-i-1); return root; }