【重点】【二叉树】105.从前序与中序遍历序列构造二叉树-CSDN博客

本文链接：https://blog.csdn.net/Allenlzcoder/article/details/134916735

博客围绕从前序与中序遍历序列构造二叉树展开，强调要牢记索引更新细节，给出了Python和Java两种语言的实现，采用递归构造的方法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目
重点牢记索引更新细节！！！

Python

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

        return root

Java

法1：递归构造

class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        TreeNode root = build(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
        return root;
    }

    public TreeNode build(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd) {
        if (preStart > preEnd) {
            return null;
        }
        if (preStart == preEnd) {
            return new TreeNode(preorder[preStart]);
        }
        TreeNode root = new TreeNode(preorder[preStart]);
        int valInx = -1;
        for (int i = inStart; i <= inEnd; ++i) {
            if (inorder[i] == preorder[preStart]) {
                valInx = i;
                break;
            }
        }
        int leftSize = valInx - inStart;
        TreeNode leftNode = build(preorder, preStart + 1, preStart + leftSize, inorder, inStart, valInx - 1);
        TreeNode rightNode = build(preorder, preStart + leftSize + 1, preEnd, inorder, valInx + 1, inEnd);
        root.left = leftNode;
        root.right = rightNode;
        return root;
    }
}

class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        return myBuild(preorder, inorder, 0, preorder.length - 1, 0, inorder.length - 1);
    }

    public TreeNode myBuild(int[] preorder, int[] inorder, 
                            int preStart, int preEnd, int inStart, int inEnd) {
        if (preStart > preEnd) {
            return null;
        } else if (preStart == preEnd) {
            return new TreeNode(preorder[preStart]);
        }
        int rootVal = preorder[preStart], rootInxOfInorder = inStart;
        for (int i = inStart; i <= inEnd; ++i) {
            if (inorder[i] == rootVal) {
                rootInxOfInorder = i;
                break;
            }
        }
        TreeNode root = new TreeNode(rootVal);
        TreeNode left = myBuild(preorder, inorder, preStart + 1, preStart + rootInxOfInorder - inStart, 
                                inStart, rootInxOfInorder - 1);
        TreeNode right = myBuild(preorder, inorder, preStart + rootInxOfInorder - inStart + 1, preEnd, 
                                rootInxOfInorder + 1, inEnd);
        root.left = left;
        root.right = right;
        return root;
    }

}