LeetCode Binary Tree Postorder Traversal

/**
 * 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) {
        vector<int> path;
        vector<TreeNode*> stack;
        if (root != NULL) {
            stack.push_back(root);
        }
        TreeNode* last = NULL;
        
        while(!stack.empty()) {
            TreeNode* cur = stack.back();
            
            if (cur->left == NULL && cur->right == NULL) {
                path.push_back(cur->val);
                stack.pop_back();
                last = cur;
                continue;
            }
            
            if (last != NULL && (cur->left == last || cur->right == last)) {
                path.push_back(cur->val);
                stack.pop_back();
                last = cur;
                continue;
            }
            
            while (cur != NULL) {
                if (cur->right != NULL) stack.push_back(cur->right);
                if (cur->left != NULL) stack.push_back(cur->left);
                cur = cur->left;
            }
        }
        
        return path;
    }
};

Note: Recursive solution is trivial, could you do it iteratively? 那就写个非递归的呗！通用算法见 LeetCode Binary Tree Inorder Traversal

第二轮：

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?

还是用老方法：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Frame {
public:
    TreeNode* node;
    int stage;
    Frame(TreeNode* n = NULL, int s = -1) : node(n), stage(s) {}
};
 
class Solution {
public:
    vector<int> postorderTraversal(TreeNode *root) {
        vector<int> res;
        
        stack<Frame> stk;
        stk.push(Frame(root, -1));
        
        while (!stk.empty()) {
            Frame& cf = stk.top();
        
            if (cf.node == NULL) {
                stk.pop();
                continue;
            }
            
            cf.stage++;
            TreeNode* node = cf.node;
            
            switch(cf.stage) {
                case 0: stk.push(Frame(node->left)); break;
                case 1: stk.push(Frame(node->right)); break;
                case 2: 
                    res.push_back(node->val);
                    stk.pop();
            }
        }
        return res;
    }
};