106. Construct Binary Tree from Inorder and Postorder Traversal「建议收藏」

全栈程序员站长

发布于 2022-08-04 12:55:20

1560

发布于 2022-08-04 12:55:20

大家好，又见面了，我是你们的朋友全栈君。

Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

难度：medium

题目：给定二叉树中序及后序遍历，构造二叉树，注意二叉树中的结点不重复。

思路；分治。

从后序遍历数组中由后向前取结点r
在中序遍历中找到r的位置，并由此结点分成两部分，继续执行1.

Runtime: 4 ms, faster than 68.02% of Java online submissions for Construct Binary Tree from Inorder and Postorder Traversal. Memory Usage: 37.6 MB, less than 42.45% of Java online submissions for Construct Binary Tree from Inorder and Postorder Traversal.

/**
 * 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[] inorder, int[] postorder) {
        if (null == postorder || postorder.length < 1) {
            return null;
        }
        
        return buildTree(inorder, 0, inorder.length - 1, postorder, postorder.length - 1);
    }
    
    public TreeNode buildTree(int[] inorder, int start, int end, int[] postorder, int idx) {
        if (start > end || idx < 0) {
            return null;
        }   
        
        TreeNode root = new TreeNode(postorder[idx]);
        int rIdx = start;
        for (; rIdx <= end; rIdx++) {
            if (inorder[rIdx] == postorder[idx]) {
                break;
            }
        }
        
        root.left = buildTree(inorder, start, rIdx - 1, postorder, idx - (end - rIdx) - 1);
        root.right = buildTree(inorder, rIdx + 1, end, postorder, idx - 1);
        
        return root;
    }
}

发布者：全栈程序员栈长，转载请注明出处：https://javaforall.cn/106985.html原文链接：https://javaforall.cn

本文参与腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。

原始发表：2022年4月2，如有侵权请联系 cloudcommunity@tencent.com 删除

二叉树