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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:由前序遍历数组和中序遍历数组构建二叉树,此题与Construct Binary Tree from Inorder and Postorder Traversal很相似。
1.从前序遍历找树的根结点,即第一个元素;
2.从中序遍历中查找根结点的位置,并记录下位置,计算出此时的长度len;
3.递归调用函数找出左右子树的根结点;
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: void buildBST(TreeNode *&root,vector<int> &preorder,int preBegin,int preEnd,vector<int> &inorder,int inBegin,int inEnd) { if(preBegin>preEnd||inBegin>inEnd) return; int pRoot=preorder[preBegin]; root=new TreeNode(pRoot); int index=0; for(int i=inBegin;i<=inEnd;i++) { if(inorder[i]==pRoot) { index=i; break; } } int len=index-inBegin; buildBST(root->left,preorder,preBegin+1,preBegin+len,inorder,inBegin,index-1); buildBST(root->right,preorder,preBegin+len+1,preEnd,inorder,index+1,inEnd); } TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) { if(preorder.size()==0||inorder.size()==0) return NULL; TreeNode *root=NULL; int preBegin=0,preEnd=preorder.size()-1; int inBegin=0,inEnd=inorder.size()-1; buildBST(root,preorder,preBegin,preEnd,inorder,inBegin,inEnd); return root; } };
Construct Binary Tree from Preorder and Inorder Traversal,码迷,mamicode.com
Construct Binary Tree from Preorder and Inorder Traversal
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原文地址:http://www.cnblogs.com/awy-blog/p/3695792.html