[LeetCode] 105. Construct Binary Tree from Preorder and Inorder Traversal

Peng_maple

于 2018-09-01 00:09:13 发布

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分类专栏： tree

本文链接：https://blog.youkuaiyun.com/Peng_maple/article/details/82263540

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本文介绍了一种通过给定的先序和中序遍历序列来构建二叉树的方法。利用递归的方式找到根节点，并划分左右子树，最终构建完整的二叉树结构。

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Construct Binary Tree from Preorder and Inorder Traversal

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

解析

递归构建左右子树即可。

代码

class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if(preorder.empty())
            return NULL;
        TreeNode* root = new TreeNode(preorder[0]);
        vector<int> leftpre;
        vector<int> rightpre;
        vector<int> leftin;
        vector<int> rightin;
        int index;
        for(int i=0;i<inorder.size();i++){
            if(inorder[i] == preorder[0]){
                index = i;
                break;
            }
        }
        for(int i=0;i<index;i++){
            leftin.push_back(inorder[i]);
            leftpre.push_back(preorder[i+1]);
        }
        for(int i=index+1;i<inorder.size();i++){
            rightin.push_back(inorder[i]);
            rightpre.push_back(preorder[i]);
        }
        root->left = buildTree(leftpre, leftin);
        root->right = buildTree(rightpre, rightin);
        return root;
    }
};