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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:利用先序遍历提供的根节点信息,在中序遍历中找到根节点,进而将序列分成{左子树,根,右子树},递归求解左右子树即可。
1 class Solution { 2 public: 3 TreeNode *buildTree( vector<int> &preorder, vector<int> &inorder ) { 4 if( preorder.empty() || preorder.size() != inorder.size() ) { return 0; } 5 return BuildSub( preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1 ); 6 } 7 private: 8 TreeNode *BuildSub( vector<int> &preorder, int preStart, int preEnd, vector<int> &inorder, int inStart, int inEnd ) { 9 if( inStart > inEnd ) { return 0; } 10 int index = 0; 11 while( inorder[inStart+index] != preorder[preStart] ) { ++index; } 12 TreeNode *treeNode = new TreeNode( preorder[preStart] ); 13 treeNode->left = BuildSub( preorder, preStart+1, preStart+index, inorder, inStart, inStart+index-1 ); 14 treeNode->right = BuildSub( preorder, preStart+index+1, preEnd, inorder, inStart+index+1, inEnd ); 15 return treeNode; 16 } 17 };
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/moderate-fish/p/3917864.html