leetCode(18):Construct Binary Tree from Preorder and Inorder (Inorder and Postorder) Traversal

标签：二叉树   leetcode   

Given preorder and inorder (Inorder and Postorder) traversal of a tree, construct the binary tree.

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

       前序和后序的特点是根结点要么在最前面、要么在最后面。已知根结点后，以根结点为界将中序遍历的结果分成两段，然后递归即可还原二叉树。有两点需要注意：首先二叉树中不能有重复的结点；其次，已知前序和后序遍历结果是无法还原二叉树的。

已知前序和中序，构建二叉树：

/**
 * 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>& preorder,int from1,int to1,
					vector<int>& inorder,int from2,int to2)
    {
    	if(from1>to1)
    		return NULL;
    	int i=from2;
    	for(;i<=to2;++i)
    	{
    		if(inorder[i]==preorder[from1])
    			break;
    	}
    	TreeNode* root=new TreeNode(inorder[i]);
    	root->left=buildTree(preorder,from1+1,from1+i-from2,inorder,from2,i-1);
    	root->right=buildTree(preorder,from1+i-from2+1,to1,inorder,i+1,to2);
    	return root;
    }
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return buildTree(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1);
    }
};

/**
 * 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,int from1,int to1,
					vector<int>& postorder,int from2,int to2)
    {
    	if(from1>to1)
    		return NULL;
    	// if(from1==to1)
    	// {
    		// TreeNode* tmp=new TreeNode(inorder[from1]);
    		// return tmp;
    	// }
    	
    	int i=from1;
    	for(;i<=to1;++i)
    	{
    		if(inorder[i]==postorder[to2])
    			break;
    	}
    	TreeNode* root=new TreeNode(inorder[i]);
    	root->left=buildTree(inorder,from1,i-1,postorder,from2,from2+i-from1-1);
    	root->right=buildTree(inorder,i+1,to1,postorder,from2+i-from1,to2-1);
    	return root;
    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        	return buildTree(inorder,0,inorder.size()-1,postorder,0,postorder.size()-1);
    }
};

leetCode(18):Construct Binary Tree from Preorder and Inorder (Inorder and Postorder) Traversal

标签：二叉树   leetcode   

原文地址：http://blog.csdn.net/walker19900515/article/details/46591235