leetcode 145. Binary Tree Postorder Traversal

相关问题

144. Binary Tree Preorder Traversal
94. Binary Tree Inorder Traversal
145. Binary Tree Postorder Traversal

Discription

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

For example:
Given binary tree {1,#,2,3},
1
  \
    2
  /
3
return [1,2,3].

思路

方法一

树的后序遍历比较复杂
先遍历左子树，第一次返回根节点；
再遍历右子树，第二次返回根节点；

所以需要变量记录当前状态，是在访问左子树还是在访问右子树。

时间复杂度：?
空间复杂度：?

代码

代码

/** * Definition for a binary tree node. * 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;        vector<TreeNode*> sta;        bool forward = true;                //访问左子树        TreeNode *preVisited = NULL;        if (root) sta.push_back(root);        while (!sta.empty())        {            while (sta.back()->left && forward) sta.push_back(sta.back()->left);            forward = false;            if (!sta.back()->right || sta.back()->right == preVisited)            {                // 右节点已经访问过了                res.push_back(sta.back()->val);                preVisited = sta.back();                sta.pop_back();            }            else            {                sta.push_back(sta.back()->right);                forward = true;            }        }        return res;    }};