leetcode 二叉树后续遍历的递归和非递归实现

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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].

/** * 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> path;        postOrder(root, path);        return path;    }    /*    void postOrder(TreeNode* root, vector<int> &path)    {        //递归写法        if (root)        {            postOrder(root->left, path);            postOrder(root->right, path);            path.push_back(root->val);        }    }    */    void postOrder(TreeNode* root, vector<int> &path)    {        //非递归写法        stack<TreeNode *> TreeNodeStack;        TreeNode *plastvisit = NULL; //记录结点是否访问过        while (root != NULL || !TreeNodeStack.empty())        {            while (root != NULL)            {                TreeNodeStack.push(root);                root = root->left;            }            root = TreeNodeStack.top();            if (root->right == NULL || plastvisit == root->right)            {                path.push_back(root->val);                plastvisit = root;                TreeNodeStack.pop();                root = NULL;            }            else                root = root->right;        }    }};

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