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

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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.

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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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 = None

class Solution:
    def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
        if len(preorder)==0:
            return None
        root=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;
    }
}

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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;
}