LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal

最新推荐文章于 2024-01-17 14:50:43 发布

Allenlzcoder

最新推荐文章于 2024-01-17 14:50:43 发布

阅读量220

点赞数

分类专栏： LeetCode练习题文章标签： LeetCode 105. Construct Binary

LeetCode练习题专栏收录该内容

168 篇文章

订阅专栏

本文介绍了一种通过前序遍历和中序遍历构建二叉树的方法。利用递归策略，首先找到根节点，在中序遍历中确定左右子树的范围，然后递归地构建左右子树。

LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal

本博客转载自：http://www.cnblogs.com/grandyang/p/4296500.html
Solution1：
递归，查好数是关键
/**
* 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 &preorder, vector &inorder) {
return my_buildTree(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1);
}
TreeNode *my_buildTree(vector &preorder, int pLeft, int pRight, vector &inorder, int iLeft, int iRight) {
if (pLeft > pRight || iLeft > iRight) return NULL;
int i = 0;
for (i = iLeft; i <= iRight; ++i) {
if (preorder[pLeft] == inorder[i]) break;
}
TreeNode *cur = new TreeNode(preorder[pLeft]);
cur->left = my_buildTree(preorder, pLeft + 1, pLeft + i - iLeft, inorder, iLeft, i - 1);//i - iLeft的值是当前inorder中属于右子树元素的个数
cur->right = my_buildTree(preorder, pLeft + i - iLeft + 1, pRight, inorder, i + 1, iRight);
return cur;
}
};