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Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
思路:递归。主要是注意调用时起始和终止的index。
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 *buildTree(vector<int> &preorder, vector<int> &inorder) { 13 TreeNode* root=process(preorder, 0,preorder.size(),inorder,0,inorder.size()); 14 return root; 15 } 16 TreeNode* process(vector<int> &preorder, int s1, int e1, vector<int> &inorder, int start, int end) 17 { 18 if(start>=end) 19 { 20 return NULL; 21 } 22 int root_val=preorder[s1]; 23 int inorder_index=start; 24 while(inorder_index<end) 25 { 26 if(inorder[inorder_index]==root_val) 27 break; 28 else inorder_index++; 29 } 30 int count=inorder_index-start;//做子树节点数量 31 TreeNode* left=process(preorder, s1+1,s1+count+1,inorder,start,inorder_index); 32 TreeNode* right=process(preorder,s1+count+1, e1,inorder, inorder_index+1,end); 33 TreeNode* root=new TreeNode(root_val); 34 root->left=left; 35 root->right=right; 36 return root; 37 } 38 };
Construct Binary Tree from Preorder and Inorder Traversal,布布扣,bubuko.com
Construct Binary Tree from Preorder and Inorder Traversal
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原文地址:http://www.cnblogs.com/hicandyman/p/3833414.html