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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?
经典题目，求二叉树的后序遍历的非递归方法，跟前序，中序，层序一样都需要用到栈，后续的顺序是左-右-根，所以当一个节点值被取出来时，它的左右子节点要么不存在，要么已经被访问过了。具体思路可参见神网友 Yu‘s Coding Garden 的博客，代码如下：
/** * 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> res; if (!root) return res; stack<TreeNode*> s; s.push(root); TreeNode *head = root; while (!s.empty()) { TreeNode *top = s.top(); if ((!top->left && !top->right) || top->left == head || top->right == head) { res.push_back(top->val); s.pop(); head = top; } else { if (top->right) s.push(top->right); if (top->left) s.push(top->left); } } return res; } };
[LeetCode] Binary Tree Postorder Traversal 二叉树的后序遍历
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原文地址：http://www.cnblogs.com/grandyang/p/4251757.html
