题目描述:(链接)
Given preorder and inorder traversal of a tree, construct the binary tree.
解题思路:
1 /** 2 * Definition for a binary tree node. 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* buildTree(vector<int>& preorder, vector<int>& inorder) { 13 return buildTree(begin(preorder), end(preorder), begin(inorder), end(inorder)); 14 } 15 16 private: 17 TreeNode *buildTree(vector<int>::iterator p_first, vector<int>::iterator p_last, 18 vector<int>::iterator i_first, vector<int>::iterator i_last) { 19 if (p_first == p_last) return nullptr; 20 if (i_first == i_last) return nullptr; 21 22 auto root = new TreeNode(*p_first); 23 auto inRootPos = find(i_first, i_last, *p_first); 24 auto leftSize = distance(i_first, inRootPos); 25 26 root->left = buildTree(next(p_first), next(p_first, leftSize + 1), i_first, next(i_first, leftSize)); 27 root->right = buildTree(next(p_first, leftSize + 1), p_last, next(inRootPos), i_last); 28 29 return root; 30 } 31 };