LeetCode: 105. Construct Binary Tree from Preorder and Inorder Traversal

题目描述

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

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

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]
Return the following binary tree:

    3
   / \   9  20
    /  \    15   7

题目大意: 根据一个二叉树的先序遍历序列和中序遍历序列，构造出该二叉树。

解题思路

根据给定的先序遍历序列，可以知道根的值，然后根据该值在中序遍历中划分左右子树，直到子树为空。如给定的例子:
1. 根据先序序列，可知，根节点的值为 3。

2. 在中序序列中找到该值，就可以划分出左子树和右子树。

划分的子树:

3. 同理， 对左子树做同样的操作:

4. 对右子树做同样的操作:

AC 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 
{
private:
    // 根据先序遍历序列的 [preBeg, preBeg+size) 和 中序遍历的 [inBeg, inBeg+size) 构建二叉树的子树
    TreeNode* buildTree(vector<int>& preorder, int preBeg, 
                              vector<int>& inorder, int inBeg, int size)
    {
        if(size <= 0) return nullptr;

        TreeNode* curNode = new TreeNode(preorder[preBeg]);

        int inorderLeftTreeEnd = inBeg;
        while(inorder[inorderLeftTreeEnd] != preorder[preBeg]) ++inorderLeftTreeEnd;
        int leftTreeSize = inorderLeftTreeEnd - inBeg;
        int rightTreeSize = (inBeg + size) - inorderLeftTreeEnd - 1;

        curNode->left = buildTree(preorder, preBeg+1, inorder, inBeg, leftTreeSize);
        curNode->right = buildTree(preorder, preBeg+1+leftTreeSize, 
                                         inorder, inorderLeftTreeEnd+1, rightTreeSize);

        return curNode;
    }
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) 
    {
        return buildTree(preorder, 0, inorder, 0, preorder.size());
    }
};