LeetCode-105：Construct Binary Tree from Preorder and Inorder Traversal (利用先序和中序遍历构建二叉树) -- medium

Question

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:

• You may assume that duplicates do not exist in the tree.

问题解析：

给定一棵树的先序和中序遍历数组，构建该二叉树。

Answer

Solution 1：

数据结构，递归调用。

与LeetCode-106：Construct Binary Tree from Inorder and Postorder Traversal的思想是相同的。

• 利用先序遍历的最先的一个元素为子树的根结点，该根结点在中序遍历的位置为左右子树的分割点。
• 需要注意左右子树的边界。

/** * 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 == null || inorder == null || preorder.length != inorder.length) return null;        Map<Integer, Integer> inMap = new HashMap<Integer, Integer>();        for(int i = 0; i < inorder.length; i++) {            inMap.put(inorder[i], i);        }        TreeNode root = buildTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1, inMap);        return root;    }    public TreeNode buildTree(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd, Map<Integer, Integer> inMap) {        if(preStart > preEnd || inStart > inEnd) return null;        TreeNode root = new TreeNode(preorder[preStart]);        int inRoot = inMap.get(root.val);        int numsLeft = inRoot - inStart;        root.left = buildTree(preorder, preStart + 1, preStart + numsLeft, inorder, inStart, inRoot - 1, inMap);        root.right = buildTree(preorder, preStart + numsLeft + 1, preEnd, inorder, inRoot + 1, inEnd, inMap);        return root;    }}

• 时间复杂度：O(lgn)，空间复杂度：O(n)

Solution 2：

python 解法，更加简洁。

• 以preorder从前到后一直保存的是子树的根结点，所以取根结点直接pop(0)，先建立左子树，再建立右子树。
• 更加清晰简洁。

# 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, inorder):        """        :type preorder: List[int]        :type inorder: List[int]        :rtype: TreeNode        """        if not preorder or not inorder:            return None        root = TreeNode(preorder.pop(0))        preorderindex = inorder.index(root.val)        root.left = self.buildTree(preorder, inorder[:preorderindex])        root.right = self.buildTree(preorder, inorder[preorderindex+1:])        return root