[Leetcode] Construct Binary Tree from Inorder and Postorder Traversal
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描述
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
给定二叉树的后序遍历以及中序遍历,要求构造该二叉树。
分析
这是一道典型的分而治之的问题。
二叉树的后序遍历的顺序是:左子数-右子树-节点,中序遍历的顺序是:左子数-节点-右子树,因此,通过后序遍历以及中序遍历,可以还原出原来的二叉树。
具体来说,后序遍历的最后一个元素表示的就是根节点,那么找到根节点之后,在中序遍历中,位于根节点左边的元素便属于左子树,位于根节点右边的元素便属于右子树,于是将问题转化成了相同的两个子问题。问题的平凡形式是某边的子树为空的情况。
相似的问题 Construct Binary Tree from Preorder and Inorder Traversal 。
代码
/** * 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) { return buildTree(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1); } TreeNode* buildTree(vector<int>& in, int lo1, int hi1, vector<int>& post, int lo2, int hi2) { if (lo1 > hi1 || lo2 > hi2) return NULL; TreeNode* node = new TreeNode(post[hi2]); int i = 0; for(i = lo1 ;i < in.size(); i++) { if (in[i] == post[hi2]) break; } node->left = buildTree(in, lo1, i - 1, post, lo2, lo2 + i - lo1 - 1); node->right = buildTree(in, i + 1, hi1, post, hi2 - hi1 + i, hi2 - 1); return node; }};
相关问题
Construct Binary Tree from Preorder and Inorder Traversal
这里要注意的是,给定二叉树的前序遍历以及中序遍历,或者是后序遍历以及中序遍历,都可以还原出原来的二叉树,但是如果给的是前序遍历以及后序遍历,得到的树不唯一。
这里举个简单的例子:
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