LEETCODE: Construct Binary Tree from Preorder and Inorder Traversal

原创于 2014-12-30 22:37:50 发布 · 361 阅读

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#LEETCODE #算法 #二叉树 #递归 #遍历

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本文介绍了一种使用前序遍历和中序遍历构建二叉树的递归方法。通过迭代找到根节点，并划分左右子树，最终构建完整的二叉树。

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

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

只会递归的方法。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *buildTree(vector<int>::iterator prebegin, vector<int>::iterator preend, 
                        vector<int>::iterator inbegin, vector<int>::iterator inend) {
        if(prebegin == preend) return NULL;
        TreeNode *root = new TreeNode(*prebegin);
        auto pos = find(inbegin, inend, *prebegin);
        int length = pos - inbegin;
        root->left = buildTree(prebegin + 1, prebegin + length + 1, inbegin, inbegin + length);
        root->right = buildTree(prebegin + length + 1, preend, inbegin + length + 1, inend);  
        return root;
    }
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        return buildTree(preorder.begin(), preorder.end(), inorder.begin(), inorder.end());
    }
};