LeetCode 0105. Construct Binary Tree from Preorder and Inorder Traversal【二叉树】

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

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根据前序序列和中序序列重建二叉树

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思路1

代码1

/**
 * 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:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return build(preorder, inorder, 0, preorder.size()-1, 0, preorder.size()-1);
    }
    
    TreeNode* build(vector<int>& preorder, vector<int>& inorder, int pl, int pr, int il, int ir){
        if(pl > pr) return NULL;
        TreeNode* node = new TreeNode(preorder[pl]);
        int k = 0;
        for(int i = il; i <= ir; i++)
            if(inorder[i] == node->val)
            {
                k = i;
                break;
            }
        node->left = build(preorder, inorder, pl + 1, pl + k - il, il, k - 1);
        node->right = build(preorder, inorder, pl + k - il + 1, pr, k + 1, ir);
        return node;
    }
};