【leetcode刷题】T127-二叉树的后序遍历

示例:
输入: [1,null,2,3]  
   1
    \
     2
    /
   3 
输出: [3,2,1]
进阶: 递归算法很简单，你可以通过迭代算法完成吗？

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def postorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        res = []
        ls = [root]
        visit = [0]
        while len(ls) > 0:
            # 未访问过，先访问左子树
            if visit[-1] == 0:
                visit[-1] = 1
                if ls[-1].left:
                    ls.append(ls[-1].left)
                    visit.append(0)
            # 访问右子树
            elif visit[-1] == 1:
                visit[-1] = 2
                if ls[-1].right:
                    ls.append(ls[-1].right)
                    visit.append(0)
            # 访问根节点
            else:
                res.append(ls[-1].val)
                ls.pop()
                visit.pop()
        return res

/**
 * Definition for a binary tree node.
 * 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) {
        vector<int> res;
        if(!root)
            return res;
        stack<TreeNode*> ls;
        stack<int> visit;
        ls.push(root);
        visit.push(0);
        while(ls.size() > 0){
            // 未访问，访问左子树
            if(visit.top() == 0){
                visit.top() = 1;
                if(ls.top()->left){
                    ls.push(ls.top()->left);
                    visit.push(0);
                }
            }else{
                // 访问过一次，则访问右子树
                if(visit.top() == 1){
                    visit.top() = 2;
                    if(ls.top()->right){
                        ls.push(ls.top()->right);
                        visit.push(0);
                    }
                }else{
                    // 访问过两次，弹出元素
                    res.push_back(ls.top()->val);
                    visit.pop();
                    ls.pop();
                }
            }
        }
        return res;
    }
};