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.
可与Construct Binary Tree from Inorder and Postorder Traversal对照来看。
前序遍历Preorder的第一个节点为根,由于没有重复值,可使用根将中序Inorder划分为左右子树。
递归下去即可。
/** * 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 *buildTree(vector<int> &preorder, vector<int> &inorder) { return Helper(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1); } TreeNode* Helper(vector<int> &preorder, int begin1, int end1, vector<int> &inorder, int begin2, int end2) { if(begin1 > end1) return NULL; else if(begin1 == end1) return new TreeNode(preorder[begin1]); TreeNode* root = new TreeNode(preorder[begin1]); int i = begin2; for(; i <= end2; i ++) { if(inorder[i] == preorder[begin1]) break; } //inorder[i] is the root int leftlen = i-begin2; //preorder[begin1] is root //inorder[begin2+leftlen] is root root->left = Helper(preorder, begin1+1, begin1+leftlen, inorder, begin2, begin2+leftlen-1); root->right = Helper(preorder, begin1+leftlen+1, end1, inorder, begin2+leftlen+1, end2); return root; } };