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.

Example 1:

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

题目翻译

给定一个树的前序和中序遍历,构造二叉树。

示例1:

null

注意: 你可以假设树中不存在重复项

解题方案

标签: Depth-first Search

思路:

  • 依照题目提示,我们假设树中不存在重复项
  • 那么我们可以通过前序遍历告诉我们根节点,在中序遍历上分开左子树和右子树
  • 每一步都寻找根节点,分开左右子树,建立树就可以了

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:
     typedef vector<int>::iterator Iter;
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
       return createTree(preorder.begin(),preorder.end(),inorder.begin(),inorder.end());
    }
    TreeNode* createTree(Iter preBegin,Iter preEnd,Iter inBegin,Iter inEnd){
        if(inBegin == inEnd) return NULL;
        TreeNode* root = new TreeNode(*preBegin);
        Iter inRoot = find(inBegin,inEnd,*preBegin);
        root->left = createTree(preBegin + 1 ,preBegin + 1 + (inRoot-inBegin),inBegin,inRoot);
        root->right = createTree(preBegin + 1 + (inRoot-inBegin),preEnd,inRoot + 1,inEnd);
        return root;
    }
};

参考资料

  • 作者:段静

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