第72题 Construct Binary Tree from Inorder and Postorder Traversal

最新推荐文章于 2025-12-19 15:39:34 发布

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该博客介绍了如何根据给定的二叉树中序和后序遍历结果，利用递归算法重建原始二叉树。解决方案包括创建一个辅助函数，首先找到根节点，然后分别构建左子树和右子树，直至所有节点都被处理。

Construct Binary Tree from Inorder and Postorder Traversal

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

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param inorder: A list of integers that inorder traversal of a tree
     * @param postorder: A list of integers that postorder traversal of a tree
     * @return: Root of a tree
     */
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        // write your code here
        if(inorder == null || postorder == null || inorder.length == 0 || postorder.length == 0){
           return null ;
        }
        HashMap<Integer,Integer> map = new HashMap<>();
        for(int i=0 ; i < inorder.length; i++){
           map.put(inorder[i],i) ;
        }

        return helper(inorder,postorder,0,inorder.length-1,0,postorder.length-1,map);

    }
    public TreeNode helper(int[] inorder , int[] postorder, int inL ,int inR, int postL, int postR,HashMap<Integer,Integer> map ){
       if(inL > inR || postL > postR){
          return null ;
       }
       TreeNode root = new TreeNode(postorder[postR]);
       int index = map.get(root.val);
       root.left = helper(inorder,postorder,inL,index-1,postL,postL+(index-inL)-1,map);
       root.right = helper(inorder,postorder,index+1,inR,postR-(inR-index),postR-1,map);
       return root ;
    }
}