1、
Construct Binary Tree from Inorder and Postorder Traversal
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
代码:
class Solution { public: TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) { TreeNode* root = NULL; int len1 = inorder.size(); int len2 = postorder.size(); if(len1<1 || len2<1 || len1!=len2){ return root; } return buildTreeCore(inorder,0,len1-1,postorder,0,len2-1); } TreeNode* buildTreeCore(vector<int>&inorder, int startIndex1, int endIndex1, vector<int>&postorder, int startIndex2, int endIndex2){ if(endIndex1 < startIndex1 ){ return NULL; } TreeNode* root = new TreeNode(postorder[endIndex2]); int index = 0; for(int i=startIndex1; i<=endIndex1; ++i){ if(inorder[i] == postorder[endIndex2]){ index = i; break; } } int leftLen = index-startIndex1; TreeNode* leftNode = NULL; TreeNode* rightNode = NULL; leftNode = buildTreeCore(inorder,startIndex1,index-1,postorder,startIndex2,startIndex2+leftLen-1); rightNode = buildTreeCore(inorder,index+1,endIndex1,postorder,startIndex2+leftLen,endIndex2-1); root->left = leftNode; root->right = rightNode; return root; } };
2、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.
class Solution { public: TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) { TreeNode* root = NULL; int len1 = inorder.size(); int len2 = preorder.size(); if(len1<1 || len2<1 || len1!=len2){ return root; } return buildTreeCore(inorder,0,len1-1,preorder,0,len2-1); } TreeNode* buildTreeCore(vector<int>&inorder, int startIndex1, int endIndex1, vector<int>&preorder, int startIndex2, int endIndex2){ if(endIndex1 < startIndex1 ){ return NULL; } TreeNode* root = new TreeNode(preorder[startIndex2]); int index = 0; for(int i=startIndex1; i<=endIndex1; ++i){ if(inorder[i] == preorder[startIndex2]){ index = i; break; } } int leftLen = index-startIndex1; TreeNode* leftNode = NULL; TreeNode* rightNode = NULL; leftNode = buildTreeCore(inorder,startIndex1,index-1,preorder,startIndex2+1,startIndex2+leftLen); rightNode = buildTreeCore(inorder,index+1,endIndex1,preorder,startIndex2+leftLen+1,endIndex2); root->left = leftNode; root->right = rightNode; return root; } };
leetcode -day23 Construct Binary Tree from Inorder and Postorder Traversal & Construct Binary Tree f
时间: 2024-12-19 18:14:40