LeetCode-106.Construct Binary Tree from Inorder and Postorder Traversal-优快云博客

本文链接：https://blog.youkuaiyun.com/zmq570235977/article/details/51764778

本文介绍了一种利用中序和后序遍历序列重构二叉树的方法。通过递归方式找到根节点，并以此划分左右子树，最终完成整棵树的构建。示例展示了如何将给定的中序和后序遍历序列转换为二叉树结构。

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

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

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

递归版

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution
{
public:
    TreeNode* helper(vector<int> &in, int inLow, int inHigh, vector<int> &post, int postLow, int postHigh)
    {
    	if (inLow > inHigh)
    		return NULL;
    	TreeNode* root = new TreeNode(post[postHigh]);
    	int rootIndex = 0;
    	for (int i = inLow; i <= inHigh; i++)
    	{
    		if (in[i] == post[postHigh])
    		{
    			rootIndex = i;
    			break;
    		}
    	}
    	root->left = helper(in, inLow, rootIndex - 1, post, postLow, rootIndex - 1 - inLow + postLow);
    	root->right = helper(in, rootIndex + 1, inHigh, post, postHigh - inHigh + rootIndex, postHigh - 1);
    	return root;
    }
    
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder)
    {
    	return helper(inorder,0, inorder.size()-1, postorder, 0, postorder.size()-1);
    }
};

测试用例

inorder:[4,7,2,1,5,3,8,6]
postorder:[7,4,2,5,8,6,3,1]

结果：

[1,2,3,4,null,5,6,null,7,null,null,8]