Construct Binary Tree from Inorder and Postorder Traversal

最新推荐文章于 2021-02-03 15:56:42 发布

brucehb

最新推荐文章于 2021-02-03 15:56:42 发布

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本文介绍了一种通过给定的二叉树中序和后序遍历序列来构造原始二叉树的方法。文章提供了一个详细的C++实现方案，并通过递归的方式重建了二叉树结构。

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:

You may assume that duplicates do not exist in the tree.

struct Node
{
	int val;
	Node(int v)
	{
		val = v;
		left = right = NULL;
	}
	Node *left;
	Node *right;
};

Node* visit(vector<int> &postorder, int postLeft, int postRight, vector<int> &inorder, int inLeft, int inRight)
{
	if (postLeft > postRight)
	{
		return NULL;
	}
	
	int index = inLeft;
	for (;index <= inRight; index++)
	{
		if (inorder[index] == postorder[postRight])
		{
			break;
		}
	}
	
	int leftLength = index - inLeft;
	Node *left = visit(postorder, postLeft, postLeft+leftLength-1, inorder, inLeft, index-1);
	Node *right = visit(postorder, postLeft+leftLength, postRight-1, inorder, index+1, inRight);
	
	Node *p = new Node(postorder[postRight]);
	p->left = left;
	p->right = right;
	
	return p;
}
Node* buildTree(vector<int> &postorder, vector<int> &inorder)
{
	if (postorder.size() < 1)
	{
		return NULL;
	}
	
	return visit(postorder, 0, postorder.size()-1, inorder, 0, inorder.size()-1);
}