LeetCode106、从中序和后序遍历序列构造二叉树

1、题目描述

https://leetcode-cn.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

2、解法

有了前一道题的经验加成，我们可以很容易的写出下面的代码:

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

    }
     //postorder后序序列，inorder后序序列,posStart,posEnd,inStart,inEnd都是对应索引
    public TreeNode build(int[]inorder,int inStart,int inEnd,int[]postorder,int posStart,int posEnd){
    
    
        if(posStart>posEnd) return null;

        //构建根节点，根节点是后序遍历的最后一个节点
        TreeNode root = new TreeNode(postorder[posEnd]);
        int root_index = 0;//1个元素
        //寻找中序遍历的根节点索引，从而获得左子树的节点个数和右子树的节点个数
        for(int i=inStart;i<=inEnd;i++){
    
    
            if(inorder[i]==root.val){
    
    
                root_index = i;
                break;
            }
        } 
    
        int left_size = root_index-inStart;//左子树节点个数 root_index-inStart+1-1=root_index-inStart
     
        //构建左子树,注意下标
        root.left = build(inorder,inStart,root_index-1,postorder,posStart,posStart+left_size-1);
        //构建右子树
        root.right = build(inorder,root_index+1,inEnd,postorder,posStart+left_size,posEnd-1);
        return root;

    }
}