LeetCode OJ 106. Construct Binary Tree from Inorder and Postorder Traversal
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LeetCode OJ 106. Construct Binary Tree from Inorder and Postorder Traversal
Description
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
解题思路
这道题跟题目105相似,只是前序遍历变成了后序遍历。根据二叉树的中序遍历和后序遍历构造二叉树。因为题目中提示了节点没有重复的值,所以我们可以判断节点的值来找出根节点。后序遍历的最后一个节点是当前子树的根节点,由此,我们可以找到根节点在中序便利中的位置root_i,把中序遍历分为两部分:这个根节点的左右子树的中序遍历。同时,后序遍历可以根据这个查找的偏移量root_i把前序遍历分为两部分:这个根节点的左右子树的后序遍历。这样就可以递归构造这个根节点的左右子树了。
Note:
vetctor容器中的assign()函数:
函数原型:
void assign(const_iterator first,const_iterator last);
void assign(size_type n,const T& x = T());
功能:
将区间[first,last)的元素赋值到当前的vector容器中,或者赋n个值为x的元素到vector容器中。注意是不含last这个元素的!这是cplusplus.com的说明链接
代码
个人github代码链接
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) { TreeNode* root; if(postorder.size() == 0 || inorder.size() == 0) return NULL; if(postorder.size() == 1) { root = new TreeNode(postorder[0]); return root; } root = new TreeNode(postorder[postorder.size()-1]); int root_i = 0; vector<int>::iterator root_itr = inorder.begin(); for(root_itr; root_itr != inorder.end(); root_itr++){ if(root->val == (*root_itr)){ break; } root_i ++; } vector<int> left_post,right_post,left_in,right_in; vector<int>::iterator itr = postorder.begin(); left_post.assign(itr, itr + root_i); right_post.assign(itr + root_i, postorder.end() - 1); itr = inorder.begin(); left_in.assign(itr,root_itr); right_in.assign(root_itr + 1, inorder.end()); root->left = buildTree(left_in, left_post); root->right = buildTree(right_in, right_post); return root; }};
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