105. Construct Binary Tree from Preorder and Inorder Traversal\106. Construct Binary Tree from Inor

105. Construct Binary Tree from Preorder and Inorder Traversal

题目描述

从二叉树的前序和中序遍历中恢复原来的二叉树。

从先序遍历中可以很轻松的找到根节点，从中序遍历中可以把左右子树分割出来。

所以这道题目使用DFS再合适不过了。

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

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

一般是假设不存在重复项，因为在递归建立子树的时候需要根据值来找到索引。重复的话，就无法找到正确的索引而使建立的树错误。

代码实现

/**
 * 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, vector<int> inorder) {
        int insize = inorder.size();
        if(!insize)  return NULL;
        TreeNode* root = new TreeNode(preorder[0]);
        int i = 0;
        for(i = 0; i < insize; i++)
            if(inorder[i] == preorder[0]) break;
        if(i)    
            root->left = buildTree(vector<int>(preorder.begin()+1, preorder.begin()+i+1), vector<int>(inorder.begin(), inorder.begin()+i));
        else root->left = NULL;    
        if(i+1 < insize) {
            root->right = buildTree(vector<int>(preorder.begin() + i + 1, preorder.end()), vector<int>(inorder.begin()+i+1, inorder.end()));
        }   
        else root->right = NULL;

        return root;
    }
};

106. Construct Binary Tree from Inorder and Postorder Traversal

题目描述

给中序和后序，给出二叉树。

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

和上一题没有区别。

代码实现

/**
 * 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) {
        int len = inorder.size(), ind = -1;
        if(!len)  return NULL;
        TreeNode *root = new TreeNode(postorder[len-1]);
        for(ind = 0; ind < len; ind++) 
            if(inorder[ind] == postorder[len - 1]) break;
        root->left = ind?buildTree(vector<int>(inorder.begin(), inorder.begin()+ind), vector<int>(postorder.begin(), postorder.begin()+ind)):NULL;
        root->right = ind+1<len?buildTree(vector<int>(inorder.begin()+ind+1, inorder.end()), vector<int>(postorder.begin()+ind, postorder.begin()+len)):NULL;
        return root;
    }
};