[LeetCode] Binary Tree Postorder Traversal
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?
解题思路:
相比于前序遍历,后续遍历思维上难度要大些,前序遍历是通过一个stack,首先压入父亲结点,然后弹出父亲结点,并输出它的value,之后压人其右儿子,左儿子即可。然而后序遍历结点的访问顺序是:左儿子 -> 右儿子 -> 自己。那么一个结点需要两种情况下才能够输出:第一,它已经是叶子结点;第二,它不是叶子结点,但是它的儿子已经输出过。那么基于此我们只需要记录一下当前输出的结点即可。对于一个新的结点,如果它不是叶子结点,儿子也没有访问,那么就需要将它的右儿子,左儿子压入。如果它满足输出条件,则输出它,并记录下当前输出结点。输出在stack为空时结束。
/** * 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) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. vector<int> ans; list<TreeNode*> node_list; if(root == NULL) return ans; node_list.push_front(root); TreeNode *head = root; while(!node_list.empty()) { TreeNode *cur = node_list.front(); if(cur -> right == head || cur -> left == head || ((cur -> right == NULL) && (cur -> left == NULL))) { node_list.pop_front(); ans.push_back(cur -> val); head = cur; } else { if(cur -> right != NULL) node_list.push_front(cur -> right); if(cur -> left != NULL) node_list.push_front(cur -> left); } } } };