106. Construct Binary Tree from Inorder and Postorder Traversal 迭代方法

最新推荐文章于 2022-01-01 11:35:53 发布

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最新推荐文章于 2022-01-01 11:35:53 发布

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本文解析了一个 LeetCode 上的 O(n) 时间复杂度的解决方案，该方案使用栈来构建二叉树。文章详细解释了算法背后的逻辑，并通过实例说明了如何通过后序遍历和中序遍历构建二叉树。

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Below is the O(n) solution from @hongzhi but that discuss is closed now 'cause @hongzhi says little about his code.

https://oj.leetcode.com/discuss/6334/here-is-my-o-n-solution-is-it-neat

I've modified some of and tried this code and got AC.
Just share about some comprehension about his code.

I've modified vtn(vector) to stn(stack) in that stack is probably what this algs means and needs.

What matters most is the meaning of stn.

Only nodes whoes left side hasn't been handled will be pushed into stn.

And inorder is organized as (inorder of left) root (inorder of right),

And postorder is as (postorder of left) (postorder of right) root.

So at the very begin, we only have root in stn and we check if inorder.back() == root->val and in most cases it's false(see Note 1). Then we make this node root's right sub-node and push it into stn.

Note 1: this is actually (inorder of right).back() == (postorder of right).back(), so if only there's no right subtree or the answer will always be false.

Note 2: we delete one node from postorder as we push one into stn.

Now we have [root, root's right] as stn and we check inorder.back() == stn.top()->val again.

true means inorder.back() is the root node and needs handled left case.
false means inorder.back() is the next right sub-node

So when we encounter a true, we will cache stn.top() as p and delete both nodes from inorder and stn.

Then we check inorder.size(), if there's no nodes left, it means p has no left node.

Else the next node in inorder could be p's left node or p's father which equals to the now stn.top() (remember we popped p from stn above).

If the latter happens, it means p has no left node and we need to move on to p's father(stn.top()).

If the former happens, it means p has one left node and it's postorder.back(), so we put it to p's left and delete it from the postorder and push the left node into stn 'cause it should be the next check node as the postorder is organized as above.

That's all of it. The algs just build a binary tree. :)

Inform me if there's anything vague or wrong, I'm open to any suggestions.

class Solution {
public:
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        if(inorder.size() == 0)return NULL;
        TreeNode *p;
        TreeNode *root;
        stack<TreeNode *> stn;
        
        root = new TreeNode(postorder.back()); 
        stn.push(root); 
        postorder.pop_back(); 
        
        while(true)
        {
            if(inorder.back() == stn.top()->val) 
            {
                p = stn.top();
                stn.pop(); 
                inorder.pop_back(); 
                if(inorder.size() == 0) break;
                if(stn.size() && inorder.back() == stn.top()->val)
                    continue;
                p->left = new TreeNode(postorder.back()); 
                postorder.pop_back();
                stn.push(p->left);
            }
            else 
            {
                p = new TreeNode(postorder.back());
                postorder.pop_back();
                stn.top()->right = p; 
                stn.push(p); 
            }
        }
        return root;
    }
};