Construct Binary Tree from Inorder and Postorder Traversal:由中序后序遍历构造二叉树

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最新推荐文章于 2024-05-28 09:38:31 发布

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分类专栏： Leetcode 文章标签： JAVA Leetcode 二叉树

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本文介绍了一种通过给定的中序和后序遍历序列来构造二叉树的方法。利用递归思想，根据后序遍历序列的最后一个元素确定根节点，并将其在中序遍历序列中定位，从而划分出左右子树，最终递归地构建整棵树。

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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:

思路：递归构造，因为后序遍历中最后一个节点是根节点，因此类似于先序遍历，根据后序根节点将中序分为左右两部分，递归构建左右子树。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    
public TreeNode build(int[] inorder,int inLeft,int inRight,int[] postorder,int postLeft, int postRight){
        if(inLeft > inRight ||postLeft > postRight) return null;
        TreeNode n = new TreeNode(postorder[postRight]);
        int roorIndex = -1;
        for(int i = inLeft; i <= inRight; i++){
            if(inorder[i] == postorder[postRight]){
                roorIndex = i;
                break;
            }
        }
        int lLen = roorIndex - inLeft;
        int RLen = inRight - roorIndex;
        n.right = build(inorder,roorIndex + 1,inRight,postorder,postRight - RLen,postRight - 1);
        n.left = build(inorder,inLeft,roorIndex - 1,postorder,postLeft,postLeft + lLen - 1);
        return n;
    }
    
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        return build(inorder,0,inorder.length - 1,postorder,0,postorder.length - 1);
    }
}