二叉树的后序遍历

具体方法：
构建两个结点，一为当前结点（current），另一个为上次访问结点（lastVisited），并构建堆栈用于存放结点。若结点不为空，则从此节点开始，遍历左子树，并将结点依此压栈。取当前结点为栈顶结点，如果当前节点的右子树为空，或是上次访问的结点（证明已经遍历此节点的左子树与右子树），则出栈，并将此节点的值存入vector中，中有再将当前结点赋予上次访问结点，当前结点置空。

.head

#ifndef C68_H
#define C68_H
#include<vector>
#include<iostream>
#include<stack>
using namespace std;
class TreeNode{
public:
    int val;
    TreeNode *left, *right;
    TreeNode(int val)
    {
        this->val = val;
        this->left = this->right = NULL;
    }
};
class Solution {
public:
    vector<int> postorderTraversal(TreeNode *root) {
        // write your code here
        TreeNode *current = root,*lastVisited=NULL;
        stack<TreeNode*> myStack;
        vector<int> res;
        while (current != NULL || !myStack.empty())
        {
            while (current != NULL)
            {
                myStack.push(current);
                current = current->left;
            }
            current = myStack.top();
            if (current->right == NULL || current->right == lastVisited)
            {
                myStack.pop();
                res.push_back(current->val);
                lastVisited = current;
                current = NULL;
            }
            else
            {
                current = current->right;
            }
        }
        return res;
    }
};
#endif

.cpp

#include"c68.h"
int main()
{
    TreeNode *root = new TreeNode(1);
    root->left = new TreeNode(2);
    root->right = new TreeNode(3);
    root->right->left = new TreeNode(5);
    root->right->right = new TreeNode(6);
    Solution s;
    vector<int> v = s.postorderTraversal(root);
    for (auto i : v)
        cout << i << " ";
    cout << endl;
    return 0;
}