leetcode[145] Binary Tree Postorder Traversal

实现后序遍历

递归：

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

非递归：

非递归使用栈。首先把根节点压栈，然后循环如下操作：用一个变量来记录上次访问的节点，如果当前栈顶元素左右儿子都为空 或者 上次访问的节点非空且等于栈顶节点的左儿子或右儿子，则直接访问；否则把栈顶元素的右节点和左节点依次压栈。

/**
 * Definition for binary tree
 * 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> perm;
        if (!root) return perm;
        stack<TreeNode *> sta;
        TreeNode *p = root, *pre = NULL;
        sta.push(p);
        while(!sta.empty())
        {
            p = sta.top();
            if (!p->left && !p->right)
            {
                perm.push_back(p->val);
                pre = p;
                sta.pop();
            }
            if (pre && (pre == p -> left || pre == p -> right))
            {
                perm.push_back(p -> val);
                pre = p;
                sta.pop();
            }
            else
            {
                if (p -> right)
                    sta.push(p -> right);
                if (p -> left)
                    sta.push(p -> left);
            }
        }
        return perm;
    }
};

 

再来一个比较有意思的解法，就是利用前序遍历实现后序遍历，只要在考虑前序遍历的根左右的过程改为根右左，然后将结果反转reverse一下就可以了。

/**
 * Definition for binary tree
 * 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> perm;
        stack<TreeNode *> sta;
        TreeNode *p = root;
        
        while(!sta.empty() || p)
        {
            while(p)
            {
                perm.push_back(p->val);
                if (p -> left)
                    sta.push(p -> left);
                p = p -> right;
            }
            if (!sta.empty())
            {
                p = sta.top();
                sta.pop();
            }
        }
        reverse(perm.begin(), perm.end());
        return perm;
    }
};