二叉树的后序非递归遍历(双栈法)-白红宇

二叉树的后序非递归遍历(双栈法)

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发布时间：2019-05-26

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Alternative Solution:

An alternative solution is to use two stacks. Try to work it out on a piece of paper. I think it is quite magical and beautiful. You will think that it works magically, but in fact it is doing a reversed pre-order traversal. That is, the order of traversal is a node, then its right child followed by its left child. This yields post-order traversal in reversed order. Using a second stack, we could reverse it back to the correct order.

Here is how it works:

Push the root node to the first stack.

Pop a node from the first stack, and push it to the second stack.

Then push its left child followed by its right child to the first stack.

Repeat step 2) and 3) until the first stack is empty.

Once done, the second stack would have all the nodes ready to be traversed in post-order. Pop off the nodes from the second stack one by one and you’re done.

void PostOreder_NoReccursive(Bitree  *T){	stack
   
    s1, s2;	Bitree *curr; //指向当前要检查的节点	s1.push(T);	while (!s1.empty())	{		curr = s1.top();		s1.pop();		s2.push(curr);		if (curr->pLeft)			s1.push(curr->pLeft);		if (curr->pRight)			s1.push(curr->pRight);	}	while (!s2.empty())	{		printf("%c ", s2.top()->iValue);		s2.pop();	}}

Complexity Analysis:

The two-stack solution takes up more space compared to the first solution using one stack. In fact, the first solution has a space complexity of

h), where

h is the maximum height of the tree. The two-stack solution however, has a space complexity of

n), where

n is the total number of nodes.

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