LeetCode 105 Construct Binary Tree from Preorder and Inorder Traversal-前序中序遍历构造二叉树-Python和Java递归解法

题目地址：Construct Binary Tree from Preorder and Inorder Traversal - LeetCode

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

python递归解法：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:if len(preorder)==0:return Noneroot=TreeNode(preorder[0])middle=inorder.index(preorder[0])root.left=self.buildTree(preorder[1:middle+1],inorder[:middle])root.right=self.buildTree(preorder[middle+1:],inorder[middle+1:])return root

java递归解法：

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/class Solution {public TreeNode buildTree(int[] preorder, int[] inorder) {if (preorder.length == 0) {return null;}TreeNode root = new TreeNode(preorder[0]);int middle = 0;for (int i = 0; i < preorder.length; i++) {if (inorder[i] == root.val) {middle = i;break;}}root.left = buildTree(Arrays.copyOfRange(preorder, 1, middle + 1),Arrays.copyOfRange(inorder, 0, middle));root.right = buildTree(Arrays.copyOfRange(preorder, middle + 1, preorder.length),Arrays.copyOfRange(inorder, middle + 1, preorder.length));return root;}
}

Java迭代做法：

public TreeNode buildTree(int[] preorder, int[] inorder) {if (preorder.length == 0) return null;Stack<TreeNode> s = new Stack<>();TreeNode root = new TreeNode(preorder[0]), cur = root;for (int i = 1, j = 0; i < preorder.length; i++) {if (cur.val != inorder[j]) {cur.left = new TreeNode(preorder[i]);s.push(cur);cur = cur.left;} else {j++;while (!s.empty() && s.peek().val == inorder[j]) {cur = s.pop();j++;}cur = cur.right = new TreeNode(preorder[i]);}}return root;
}

LeetCode 105 Construct Binary Tree from Preorder and Inorder Traversal-前序中序遍历构造二叉树-Python和Java递归解法相关推荐

LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal 由前序和中序遍历建立二叉树 C++...
LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal 由前序和中序遍历建立二叉树 C++ Given preo ...
LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal
LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal 本博客转载自:http://www.cnblogs.co ...
[LeetCode]*105.Construct Binary Tree from Preorder and Inorder Traversal
题目 Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume t ...
LeetCode: 105. Construct Binary Tree from Preorder and Inorder Traversal
题目 Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume t ...
[LeetCode] Construct Binary Tree from Preorder and Inorder Traversal 由先序和中序遍历建立二叉树...
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that ...
leetcode题解:Construct Binary Tree from Preorder and Inorder Traversal (根据前序和中序遍历构造二叉树)...
题目: Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume ...
LeetCode OJ - Construct Binary Tree from Preorder and Inorder Traversal
题目: Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume ...
LeetCode 889. Construct Binary Tree from Preorder and Postorder Traversal
原题链接在这里:https://leetcode.com/problems/construct-binary-tree-from-preorder-and-postorder-traversal/ 题 ...
[leetcode] Construct Binary Tree from Preorder and Inorder Traversal
Construct Binary Tree from Preorder and Inorder Traversal Given preorder and inorder traversal of a ...

LeetCode 105 Construct Binary Tree from Preorder and Inorder Traversal-前序中序遍历构造二叉树-Python和Java递归解法

LeetCode 105 Construct Binary Tree from Preorder and Inorder Traversal-前序中序遍历构造二叉树-Python和Java递归解法相关推荐

最新文章

热门文章