LeetCode 145.Binary Tree Postorder Traversal (二叉树的后序遍历)

悠悠 2022-05-08 13:33 181阅读 0赞

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]

AC C++ Solution:

堆栈迭代解法：

/**
     * 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> nodes;
            stack<TreeNode*> todo;
            TreeNode* cur = root;
            TreeNode* last = NULL;
            while (cur || !todo.empty()) {
                if (cur) {
                    todo.push(cur);
                    cur = cur -> left;
                } else {
                    TreeNode* top = todo.top();
                    if (top -> right && (last != top -> right)) {
                        cur = top -> right;
                    } else {
                        nodes.push_back(top -> val);
                        last = top;
                        todo.pop();
                    }
                }
            }
            return nodes;
        }
    };

递归解法：

/**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    
    //递归
    class Solution {
    public:
        vector<int> postorderTraversal(TreeNode* root) {
            vector<int> nodes;
            postorder(root, nodes);
            return nodes;
        }
    private:
        void postorder(TreeNode* root, vector<int>& nodes) {
            if (!root) {
                return;
            }
            postorder(root -> left, nodes);
            postorder(root -> right, nodes);
            nodes.push_back(root -> val);
        }
    };

同样，递归解法一目了然，推广到其它两种遍历也很简单。