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leetcode -day23 Construct Binary Tree from Inorder and Postorder Traversal Construct Binary Tree f

發(fā)布時間:2025/3/21 编程问答 27 豆豆
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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.

代碼:

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;} };


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