TreeNode* constructFromPrePost(vector<int>& pre, vector<int>&post) {returnbuild(pre,0,post,0,pre.size()); } TreeNode* build(vector<int> &pre,intprel,vector<int> &post,intpostl,intN){if(N ==0)return0; TreeNode*cur =newTreeNode(pre[prel]);if(N ==1)returncur;inti =1;for(i...
就可以重建出原始的二叉树,而且正好都在 LeetCode 中有出现,其他两道分别是 [Construct Binary Tree from Inorder and Postorder Traversal](https://www.cnblogs.com/grandyang/p/4296193.html) 和 [Construct Binary Tree from Preorder and Inorder Traversal](https://www.cnblogs.com/grandyang/p/4296500...
Given two integer arrays, preorder and postorder where preorder is the preorder traversal of a binary tree of distinct values and postorder is the postorder traversal of the same tree, reconstruct and return the binary tree. If there exist multiple answers, you can return any of them. Example...
105. Construct Binary Tree from Preorder and Inorder Traversal——tree,程序员大本营,技术文章内容聚合第一站。
(preBegin == preEnd) return nullptr; //从中序遍历结果中找出根节点(根节点在前序遍历中是第一个节点) vector<int>::iterator itRoot = find(inBegin, inEnd, *preBegin); TreeNode *root = new TreeNode(*itRoot); //计算根的左子树节点个数 int leftSize = itRoot - inBegin; //在中序遍历...
LeetCode—106. Construct Binary Tree from Inorder and Postorder Traversal,程序员大本营,技术文章内容聚合第一站。
Given preorder and inorder traversal of a tree, construct the binary tree. 本题就是根据前序遍历和中序遍历的结果还原一个二叉树。 题意很简答,就是递归实现,直接参考代码吧。 查询index部分可以使用Map做查询,建议和下一道题 leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal 中...
4. Construct Binary Tree from Inorder and Postorder Traversal 这里的解题思路是我们知道postorder的最后一个元素一定是根节点,按照这个特性可以找出根节点在inorder中的位置,从该位置分开,左右两边分别就是左右子树。 class Solution: def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNo...
Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 思路: 这题的思路与 105 Construct Binary Tree from Preorder and Inorder Traversal 基本相同。
root.left = helper(inStart, inMid - 1, preorder, inorder); // 先序序列 1(234)(567) 中5是右子树的根 root.right = helper(inMid + 1, inEnd, preorder, inorder); return root; } } Construct Binary Tree from Inorder and Postorder Traversal ...