我们都知道,已知中序和后序的序列是可以唯一确定一个二叉树的. 初始化时候二叉树为:================== 中序遍历序列,           ======O=========== 后序遍历序列,           =================O 红色部分是左子树,黑色部分是右子树,O是根节点 如上图所示,O是根节点,由后序遍历可知, 根据这个O可以把找到其在中序遍历当中的位置,进而,知道当前这个根节点O的左子树的前序遍历和中序遍历序列的范围. 以及右子树的前序遍历和中序遍历…

题目: Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 说明: 1)实现与根据先序和中序遍历构造二叉树相似,题目参考请进 算法思想 中序序列:C.B.E.D.F.A.H.G.J.I   后序序列:C.E.F.D.B.H.J.I.G.A   递归思路: 根据后序遍历的特点,…

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,可参见我之前的一篇博客Convert Sorted Array to Bin…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given inorder = [,,,,] postorder = [,,,,] Return the following binary tree: / \ / \ 中序.后序遍历得到二叉树,可以…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree: 3 / \ 9 2…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree: 3 / \ 9 2…

给定一棵树的中序遍历与后序遍历,依据此构造二叉树.注意:你可以假设树中没有重复的元素.例如,给出中序遍历 = [9,3,15,20,7]后序遍历 = [9,15,7,20,3]返回如下的二叉树:    3   / \  9  20    /  \   15   7详见:https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description/ Java实现: /** *…

［抄题］: Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree: 3 /…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree: 3 / \ 9 2…

根据中序和后续遍历构建二叉树. /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public TreeNode buildTree(int[] inorder, int[] postorder) { if(ino…