[LeetCode]Binary Tree Postorder Traversal

Q：Given a binary tree, return the postorder traversal of its nodes‘ values.

Note：Recursive solution is trivial, could you do it iteratively?

题目的意思就是不用递归求二叉树的后序遍历。

后续遍历的递归方式很简单，首先遍历左子树，然后遍历右子树，最后遍历根节点；遍历左右子树的时候，仍然是先遍历左子树，然后再遍历右子树，最后遍历根节点。非递归的时候，就额外使用递归中隐含的栈，重点注意的是，遍历右子树的时候需要做个标记，表示右子树是否已经被访问过。

下面贴上代表，便于测试正确性，补上二叉树的先序建立算法：

#include <iostream>
#include <vector>
#include <stack>
using namespace std;
struct TreeNode{
	int val;
	TreeNode* left;
	TreeNode* right;
	TreeNode(int x) :val(x), left(NULL), right(NULL){}
};
class Solution{
public:
	TreeNode* root;
public:
	TreeNode* createPre(TreeNode* bt){
		int x;
		cin >> x;
		if (x == 0)
			bt = NULL;
		else{
			bt = new TreeNode(x);
			bt->left = createPre(bt->left);
			bt->right = createPre(bt->right);
		}
		return bt;
	}

	void create(){
		root = createPre(root);
	}
	vector<int> postorderTraversal(TreeNode *root) {
		vector<int> order;
		if (root == NULL)
			return order;
		stack<TreeNode*> s;
		TreeNode* visited = NULL;
		while (root != NULL || !s.empty()){
			while (root != NULL){
				s.push(root);
				root = root->left;
			}
			root = s.top();
			if (root->right == NULL || root->right==visited){
				order.push_back(root->val);
				s.pop();
				visited = root;
				root = NULL;
			}
			else{
				root = root->right;
			}
		}
		return order;
	}

};

测试结果如下：

[LeetCode]Binary Tree Postorder Traversal

原文：http://blog.csdn.net/kaitankedemao/article/details/43340151