Binary Tree Postorder Traversal-二叉树的后序遍历

原题：

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

=>给定一个二叉树，给出后序遍历的所有节点值

For example:
=>例如

Given binary tree {1,#,2,3},

=>给定一个二叉树{1，#，2,3}

   1
    \
     2
    /
   3

return [3,2,1].

=>返回[3,2,1]

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

=》注意：递归的实现很普通，能不能不用递归实现？

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        
    }
};

晓东解析：

其实之前晓东已经和大家分析了前序遍历的实现，后序遍历相对而言要复杂一点，关键的思想在于，我如何知道我已经遍历了右子树，所以，需要增加一个类似的标志位，来标志右子树的遍历。因此，我们先遍历到左子树，然后遍历右子树，加上一个标志位，在右子树已经遍历的情况下，就可以把该节点加入了。还有一点需要注意的是，在加入节点的时候才能把该节点从stack中pop出来哦。

代码实现：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        stack<TreeNode*> TreeStack;
        vector<int> result;
        TreeNode * Hasvisited;
        
        if(root == NULL) return result;
        while(root || !TreeStack.empty()){
            while(root){
                TreeStack.push(root);
                root = root->left;
            }
            root = TreeStack.top();

            if(root->right == NULL || Hasvisited == root->right){
                result.push_back(root->val);
                Hasvisited = root;
                TreeStack.pop();
                root = NULL;
            }
            else{
                root = root->right;
            }
        }
        
        return result;
        
    }
};

执行结果：

67 / 67test cases passed.	Status: Accepted
Runtime: 16 ms

希望大家有更好的算法能够提出来，不甚感谢。

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