106. Construct Binary Tree from Inorder and Postorder Traversal

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每日一题 2019 - 04 - 24

题目：

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

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inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

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2
3
4
5
  3
 / \
9  20
  /  \
 15   7

解法：

这个题给出二叉树的中序遍历和后续遍历让还原出原先的二叉树，思路大概如下：

  • 后续遍历的最后一个结点为二叉树的根节点，找出根节点，在中序遍历中根据根节点将中序遍历分为左子树与右子树
  • 在中序遍历分开后的左子树右子树中继续进行划分，同时将后续遍历的序列也进行划分

递归这个过程即可完成还原；

代码：

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/**
 * 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* buildTree(vector<int>& inorder, vector<int>& postorder) {
        return fun(inorder,postorder,0, inorder.size(), 0,postorder.size()-1);
    }
    TreeNode* fun(vector<int>& inorder, vector<int>& postorder,int in_start, int in_end, int post_start, int post_end)
    {
        if(post_start==post_end) return new TreeNode(postorder[post_end]);
        if(post_start > post_end) return NULL;
        int root_value=postorder[post_end];
        TreeNode* root=new TreeNode(root_value);
        int i=0;
        for(;i<in_end;++i)
        {
            if(inorder[i]==root_value)
                break;
        }
        root->left=fun(inorder,postorder,in_start, i-1, post_start, post_start+ i-in_start-1);
        root->right=fun(inorder,postorder,i+1, in_end, post_start+i-in_start, post_end-1);
        return root;
    }
};
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