Construct Binary Tree from Preorder and Inorder Traversal

最新推荐文章于 2025-08-09 17:19:50 发布

Vinson

最新推荐文章于 2025-08-09 17:19:50 发布

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本文介绍了一种根据前序遍历和中序遍历序列构造二叉树的方法。通过递归方式找到根节点，并根据中序遍历确定左右子树的范围，进而构建整棵树。

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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.

// construct tree from preorder and inorder
 
class Solution {
    
public:
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        if(preorder.size() == 0) return NULL;
        
        return buildTree(preorder, 0, inorder, 0, preorder.size());
    }
    
    TreeNode *buildTree(vector<int> &pre, int preStart, vector<int> &in, int inStart, int len) {
        if(preStart <  0 || preStart + len > pre.size() || inStart < 0 || inStart + len > in.size() || len < 1) {
            return NULL;
        }
        
        int rootVal = pre[preStart];
        
        int i = 0;
        while(in[i] != rootVal) ++i;
        
        TreeNode *root = new TreeNode(rootVal);
        
        int leftLen = i - inStart;
        int rightLen = len - leftLen - 1;
        
        root->left = buildTree(pre, preStart+1, in, inStart, leftLen);
        root->right = buildTree(pre, preStart+leftLen+1, in, i+1, rightLen);
        
        return root;
    }
};

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