LeetCode145：Binary Tree Postorder Traversal

题目：

Given a binary tree, return the postorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [3,2,1].

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

解题思路：

后序遍历的非递归实现与前序遍历和中序遍历的非递归不同，对于根节点，必须当其右孩子都访问完毕后才能访问，所以当根节点出栈时，要根据标志位判断是否为第一次出栈，如果是，则将其标志位置为FALSE，然后再次入栈，先访问其右孩子，当某个节点出栈时且其标志位为false，说明该节点为第二次出栈，即表示其右孩子都已处理完毕，可以访问当前节点了

实现代码：

#include <iostream>
#include <vector>
#include <stack>
using namespace std;

/**
Given a binary tree, return the postorder traversal of its nodes' values.
*/
 

struct TreeNode {
     int val;
     TreeNode *left;
     TreeNode *right;    
     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

void addNode(TreeNode* &root, int val)
{
    if(root == NULL)
    {
        TreeNode *node = new TreeNode(val);
        root = node;
    }
    else if(root->val < val)
    {
        addNode(root->right, val);
    }
    else if(root->val > val)
    {
        addNode(root->left, val);
    }
}

void printTree(TreeNode *root)
{
    if(root)
    {
        cout<<root->val<<" ";
        printTree(root->left);
        printTree(root->right);
    }
}
struct Node
{
    TreeNode *treeNode;
    bool isFirst;
};

class Solution {
public:
    vector<int> postorderTraversal(TreeNode *root) {
        vector<int> ivec;
        if(root)
        {                
            stack<Node*> istack;
            TreeNode *p = root;
            while(!istack.empty() || p)
            {
                while(p)
                {
                    Node *tmpNode = new Node();
                    tmpNode->treeNode = p;
                    tmpNode->isFirst = true;
                    istack.push(tmpNode);
                    p = p->left;
                }
                if(!istack.empty())
                {
                    Node *node = istack.top();
                    istack.pop();
                    if(node->isFirst)//第一次出栈， 
                    {
                        node->isFirst = false;
                        istack.push(node);
                        p = node->treeNode->right;
                    }
                    else//表示其右孩子都已经访问了，可以访问当前节点了 
                    {
                        ivec.push_back(node->treeNode->val);
                    }
                }
                
            }
            
        }
        return ivec;
                       
    }
};
int main(void)
{
    TreeNode *root = new TreeNode(5);
    addNode(root, 7);
    addNode(root, 3);
    addNode(root, 15);
    addNode(root, 1);
    printTree(root);
    cout<<endl;
    
    Solution solution;
    vector<int> v = solution.postorderTraversal(root);
    vector<int>::iterator iter;
    for(iter = v.begin(); iter != v.end(); ++iter)
        cout<<*iter<<" ";
    cout<<endl;
    
    return 0;
}