LeetCode 145: Binary Tree Postorder Traversal(二叉树的后序遍历，迭代法)

145. Binary Tree Postorder Traversal

   

Total Accepted: 96378   Total Submissions: 271797   Difficulty: Hard

提交网址：  https://leetcode.com/problems/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?

分析：

AC代码：

#include<iostream>
#include<vector>
#include<stack>
#include<algorithm>  // 引入 reverse函数，对vector进行反转 
using namespace std;

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(0);
        if(root==NULL) return res;
        stack<TreeNode *> st;
        TreeNode *p=root;
        while(!st.empty()|| p!=NULL)
        {
            if(p!=NULL)
            {
            st.push(p);
            res.push_back(p->val);
            p=p->right;
            }
            if(p==NULL)
            {
                p=st.top();
                p=p->left;
                st.pop();
            }
        }
        reverse(res.begin(),res.end());    // 对向量及其内部结构进行反转
     return res;   
    }
};
// 以下为测试
int main()
{
	Solution sol;
	vector<int> res;
	
	TreeNode *root = new TreeNode(1); 
    root->right = new TreeNode(2); 
    root->right->left = new TreeNode(3); 
	
	res=sol.postorderTraversal(root);
	
	for(int i:res)
		cout<<i<<" ";  // 此处为vector遍历的方法，C++11标准支持 
	return 0;
}

另一解法(非递归)：

后序遍历的麻烦在于访问过右子树之后，第二次访问的时候就必须访问根节点，而不是继续访问右子树，所以使用pre来记录上次访问的节点...

class Solution {
public:
    vector<int> postorderTraversal(TreeNode *root) {
        vector<int> ans;
        if(root == NULL) return ans;
        stack<TreeNode* > s;
        TreeNode * p = root;
        TreeNode * pre = NULL;                 //  记录上次访问的节点
        while(p != NULL || !s.empty()) {
            while(p != NULL) {
                s.push(p);
                p = p->left;
            }
            if(!s.empty()) {
                p = s.top();
                s.pop();
                if(p->right == NULL || p->right == pre) {        // 右子树为空，或者是访问过
                    ans.push_back(p->val);
                    pre = p;                         // 记录刚刚访问的右节点
                    p = NULL;
                }
                else {
                    s.push(p);
                    p = p->right;
                }
            }
        }
        return ans;
    }
};