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?
分析：用两个栈，一个栈A用于扩展子树，另一个B栈用于保存结果。入栈A的顺序是先左子树后右子树，从栈A出的元素被压入栈B。代码如下：
class Solution { public: vector<int> postorderTraversal(TreeNode *root) { vector<int> res; if(!root) return res; stack<TreeNode *> s; stack<TreeNode *> output; TreeNode * cur; s.push(root); while(!s.empty()){ cur = s.top(); output.push(cur); s.pop(); if(cur->left) s.push(cur->left); if(cur->right) s.push(cur->right); } while(!output.empty()){ cur = output.top(); res.push_back(cur->val); output.pop(); } return res; } };
Leetcode: Binary Tree Postorder Traversal
原文：http://www.cnblogs.com/Kai-Xing/p/4143854.html