Construct Binary Tree from Inorder and Postorder Traversal

最新推荐文章于 2021-02-03 15:56:42 发布

原创最新推荐文章于 2021-02-03 15:56:42 发布 · 1.3k 阅读

0 ·

CC 4.0 BY-SA版权

二叉树 (binary tree) 专栏收录该内容

34 篇文章

订阅专栏

本文详细介绍了如何通过给定的二叉树中序和后序遍历序列，构建原始二叉树的过程。通过递归算法实现，确保在不重复元素的情况下准确重建树结构。

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

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

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if (inorder == null || inorder.length == 0 || postorder == null || postorder.length == 0) {
            return null;
        }
        
        if (inorder.length == 1) {
            return new TreeNode(inorder[0]);
        }
        
        TreeNode node = new TreeNode(postorder[postorder.length - 1]);
        int index = indexOf(inorder, node.val);
        if (index >= 1) {
            node.left = buildTree(Arrays.copyOfRange(inorder, 0, index), Arrays.copyOfRange(postorder, 0, index));
        }
        if (postorder.length -1 >=  index)
            node.right = buildTree(Arrays.copyOfRange(inorder, index + 1, inorder.length), Arrays.copyOfRange(postorder, index, postorder.length - 1));
        return node;    
    }
    
    public int indexOf(int[] arr, int val) {
        if (arr == null || arr.length == 0) return -1;
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] == val) {
                return i;
            }
        }
        return -1;
    }
}