Binary Tree Postorder Traversal

Problem:

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

For example:

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

return [3, 2, 1]

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

分析：

后序遍历的访问顺序是“左右中”，先遍历左子树、右子树，后遍历根节点。思路基本与先序遍历一致，需要借助堆栈。但是根节点是最后遍历的，所以遍历左子树、右子树时候，只做压栈的操作，不将节点加入遍历序列中。将节点加入遍历序列的条件是当前判断的栈顶的元素，没有右子树，或者右子树的根节点是其前驱（代码的第14行）。将根节点访问完后才将其弹出堆栈（代码的第16行）。

class Solution {
public:
    vector<int> postorderTraversal(TreeNode *root) {
        vector<int> result;
        stack<TreeNode *> stk;
        TreeNode *p = root;
        while(p != NULL) {
            stk.push(p);
            p = p->left;
        }
        
        while(!stk.empty()) {
            TreeNode *top = stk.top();
            if(top->right == NULL || result.size() > 0 && top->right->val == result.back()) {
                result.push_back(top->val);
                stk.pop();
            } else {
                TreeNode *p = top->right;
                while(p != NULL) {
                    stk.push(p);
                    p = p->left;
                }
            }
        }
        return result;
    }
};