Given inorder and postorder traversal of a tree, construct the binary tree. Note:  You may assume that duplicates do not exist in the tree. 利用中序和后序遍历构造二叉树,要注意到后序遍历的最后一个元素是二叉树的根节点,而中序遍历中,根节点前面为左子树节点后面为右子树的节点.例如二叉树:{1,2,3,4,5,6,#}的后序遍历为4->5->2->6-&…

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…

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…

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 = [,,,,] postorder = [,,,,] Return the following binary tree: / \ / \ 中序.后序遍历得到二叉树,可以…

题意:根据二叉树的中序遍历和后序遍历恢复二叉树. 解题思路:看到树首先想到要用递归来解题.以这道题为例:如果一颗二叉树为{1,2,3,4,5,6,7},则中序遍历为{4,2,5,1,6,3,7},后序遍历为{4,5,2,6,7,3,1},我们可以反推回去.由于后序遍历的最后一个节点就是树的根.也就是root=1,然后我们在中序遍历中搜索1,可以看到中序遍历的第四个数是1,也就是root.根据中序遍历的定义,1左边的数{4,2,5}就是左子树的中序遍历,1右边的数{6,3,7}就是右子树的中序遍历…

要求:根据中序和后序遍历序列构建一棵二叉树 代码如下: struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x): val(x),left(NULL), right(NULL) {} }; typedef vector<int>::iterator Iter; TreeNode *buildTreeFromInorderPostorder(vector<int> &inorder,…

LeetCode: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.                                            …

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. Hide Tags Tree Array Depth-first Search   SOLUTION 1:…

Given inorder and postorder 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>& inorder, int in_left,int in_right, vector<int&…