【LeetCode-树】二叉树的后序遍历

题目描述

给定一个二叉树，返回它的 后序 遍历。
示例：

输入: [1,null,2,3]  
   1
    
     2
    /
   3 

输出: [3,2,1]

题目链接： https://leetcode-cn.com/problems/binary-tree-postorder-traversal/

思路1

使用递归。代码如下：

/**
 * 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) {
        if(root==nullptr) return {};

        vector<int> ans;
        postOrder(root, ans);
        return ans;
    }

    void postOrder(TreeNode* root, vector<int>& ans){
        if(root==nullptr) return;

        postOrder(root->left, ans);
        postOrder(root->right, ans);
        ans.push_back(root->val);
    }
};

• 时间复杂度：O(n)
• 空间复杂度：O(h)

思路2

使用迭代。类似于中序遍历的颜色标记法。代码如下：

/**
 * 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) {
        if(root==nullptr) return {};

        vector<int> ans;
        stack<TreeNode*> nodeStack;
        stack<int> visit;
        nodeStack.push(root); visit.push(0);
        while(!nodeStack.empty()){
            TreeNode* curNode = nodeStack.top(); nodeStack.pop();
            int hasVisit = visit.top(); visit.pop();
            if(hasVisit==0){
                nodeStack.push(curNode); visit.push(1);
                if(curNode->right!=nullptr){
                    nodeStack.push(curNode->right); visit.push(0);
                }
                if(curNode->left!=nullptr){
                    nodeStack.push(curNode->left); visit.push(0);
                }
            }else{
                ans.push_back(curNode->val);
            }
        }
        return ans;
    }
};

• 时间复杂度：O(n)
• 空间复杂度：O(h)