[LeetCode] 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.
树的递归
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 TreeNode *createTree(vector<int> &preorder, int preBeg, int preEnd, vector<int> &inorder, int inBeg, int inEnd) 13 { 14 if (preBeg > preEnd) 15 return NULL; 16 17 int root = preorder[preBeg]; 18 int index; 19 for(int i = inBeg; i <= inEnd; i++) 20 if (root == inorder[i]) 21 { 22 index = i; 23 break; 24 } 25 26 int len = index - inBeg; 27 TreeNode *left = createTree(preorder, preBeg + 1, preBeg + len, inorder, inBeg, index - 1); 28 TreeNode *right = createTree(preorder, preBeg + len + 1, preEnd, inorder, index + 1, inEnd); 29 30 TreeNode *node = new TreeNode(root); 31 32 node->left = left; 33 node->right = right; 34 35 return node; 36 } 37 38 TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) { 39 // Start typing your C/C++ solution below 40 // DO NOT write int main() function 41 if (preorder.size() == 0) 42 return NULL; 43 44 TreeNode *head = createTree(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1); 45 46 return head; 47 } 48 };