=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...
1TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {2//Start typing your C/C++ solution below3//DO NOT write int main() function4TreeNode *root =newTreeNode(0);5if(inorder.size() ==0){6returnNULL;7}8vector<int>leftInorder, leftPostorder, rightInorder, rightPostor...
inorder 和postorder共同点就是他们从右往左能够找到最右边的treenode,而preorder 和inorder共同点就是他们能够找到最左边的treenode,所以从左往右开始递归。 最后一个相似的题就是给你preorder, postorder, 让你重建tree,preorder: root, left, right. postorder: left, right, root. 这时候你依然发现可以从左...
https://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/ 解答: 跟用inorder, preorder重构树一样,唯一不同的是root在postorder的最后,preorder则在最前。 /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right...
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...
1. Problem Descriptions:Given two integer arrays inorderandpostorderwhereinorderis the inorder traversal of a binary tree andpostorderis the postorder traversal of the same tree, construct and retu…
Given inorder[1,2,3]and postorder[1,3,2], return a tree: 2 / \ 1 3 1. 2. 3. 分析: 这是非常典型的递归问题,postorder的最后一个是root,在inorder里面找出root的位置,左边部分为左子树,右边部分为右子树,稍微麻烦的部分就是确定左子树和右子树的starting index。
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...
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 ...
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...