left(NULL), right(NULL) {}8* };9*/10classSolution {11public:12TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {13if(inorder.size()!=postorder.size()||inorder.size()<1)14returnNULL;15returnbuild(inorder,
标志31if(startPostorder==endPostorder){32returnnewTreeNode(postorder[endPostorder]);33}else{34//记录根结点的在中序遍历中的位置35introotIn=startInorder;36for(inti=startInorder;i<=endInorder;i++){37if(inorder[i]==postorder[endPostorder]){38rootIn=i;39break;40}41}42//创建根结点43TreeNod...
=postorder.length)returnnull;Map<Integer,Integer>map=newHashMap<Integer,Integer>();for(inti=0;i<inorder.length;i++){map.put(inorder[i],i);}intlen=inorder.length;returnhelper(inorder,0,len-1,postorder,0,len-1,map);}publicTreeNodehelper(int...
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: easy 算法: 1. public TreeNode buildTree(int[] inorder, int[] postorder) { 2. if (inorder == null || inorder.length == 0 || postorde...
Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 翻译:给定树的中序遍历和后序遍历,构造二叉树。 注意:树中不存在重复项。 思路:本题与105. Construct Binary Tree from Preorder and Inorder Traversal类似。
Return the root of the constructed binary tree. 3.1.2 Indices as parameters: Start by defining a helper function build_tree_helper that takes 4 parameters inorder and postorder's start and end respectively . Identify the root of the binary tree using the last element of the postorder list. ...
http://www.lintcode.com/en/problem/construct-binary-tree-from-inorder-and-postorder-traversal/ 【题目解析】 本题在属于二叉树遍历的经典题目,已知二叉树的两个遍历序列构造二叉树,有如下性质: 若已知先序和中序,则可以构造出唯一的二叉树 若已知先序和后序,则可以构造出多颗不同的二叉树 ...
postorder = [9,15,7,20,3] 1. 2. Return the following binary tree: 3 / \ 9 20 / \ 15 7 1. 2. 3. 4. 5. 题解: 同105 classSolution{ public: TreeNode*buildTree(intpleft,intpright,intileft,intiright,vector<int>&postorder,vector<int>&inorder) { ...
type Index struct{index int}funcbuildTree(inorder[]int,postorder[]int)*TreeNode{iflen(inorder)==0{returnnil}//后续遍历的最后一个元素,是二叉树的root节点rootIndex:=&Index{len(postorder)-1}returnbuildTreeR(inorder,postorder,0,len(postorder)-1,rootIndex)}funcbuildTreeR(inorder[]int,postorder...
Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 这道题要求从中序和后序遍历的结果来重建原二叉树,我们知道中序的遍历顺序是左-根-右,后序的顺序是左-右-根,对于这种树的重建一般都是采用递归来做,可参见...