LeetCode 145. Binary Tree Postorder Traversal

原题链接在这里：https://leetcode.com/problems/binary-tree-postorder-traversal/

题目：

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?

题解：

类似Binary Tree Inorder Traversal 和 Binary Tree Preorder Traversal.

Method 1:  Recursion

Time Complexity: O(n). Space: O(logn).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    //Method 1: Recursion
    public List<Integer> postorderTraversal(TreeNode root) {
        //Recursion
        List<Integer> ls = new ArrayList<Integer>();
        helper(root,ls);
        
        return ls;
    }
    private void helper(TreeNode root,List<Integer> ls){
        if(root == null){
            return;
        }
        helper(root.left,ls);
        helper(root.right,ls);
        ls.add(root.val);
    }
}

AC C++:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        dfs(root, res);
        return res;
    }

private:
    void dfs(TreeNode* root, vector<int>& res){
        if(!root){
            return;
        }

        dfs(root->left, res);
        dfs(root->right, res);
        res.push_back(root->val);
    }
};

AC Python:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        res = []
        self.dfs(root, res)
        return res
    def dfs(self, root, res):
        if not root:
            return
        self.dfs(root.left, res)
        self.dfs(root.right, res)
        res.append(root.val)

Method 2: 

Pretty much like method 2 of Binary Tree Preorder Traversal.

The difference is that when pushing into stack, push from left to right.

Time Complexity: O(n).

Space: O(logn).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        
        Stack<TreeNode> stk = new Stack<>();
        stk.push(root);
        while(!stk.isEmpty()){
            TreeNode cur = stk.pop();
            res.add(cur.val);
            if(cur.left != null){
                stk.push(cur.left);
            }
            
            if(cur.right != null){
                stk.push(cur.right);
            }
        }
        
        Collections.reverse(res);
        return res;
    }
}

Method 3: Iteration + Stack

和Preorder, Inorder 类似.

当栈顶点右侧为空 或者 右侧已经traverse 过了, 就可以把栈顶的点加入res中. 节点pre用来记载刚刚加入res的点, 可以用来判定右侧是否已经traverse过了.

top.right == pre 说明right已经加入res里了.

Time Complexity: O(n), 每个点访问了一次. Space:O(logn).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        Stack<TreeNode> stk = new Stack<TreeNode>();
        TreeNode pre = null;
        while(root != null || !stk.isEmpty()){
            if(root != null){
                stk.push(root);
                root = root.left;
            }else{
                TreeNode top = stk.peek();
                if(top.right == null || top.right == pre){
                    res.add(top.val);
                    stk.pop();
                    pre = top;
                }else{
                    root = top.right;
                }
            }
        }
        return res;
    }
}

AC C++:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> stk;
        TreeNode* pre;
        while(root || !stk.empty()){
            if(root){
                stk.push(root);
                root = root->left;
            }else{
                TreeNode* top = stk.top();
                if(!top->right || top->right == pre){
                    res.push_back(top->val);
                    stk.pop();
                    pre = top;
                }else{
                    root = top->right;
                }
            }
        }

        return res;
    }
};

AC Python:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        res, stk = [], []
        pre = None
        while root or stk:
            if root:
                stk.append(root)
                root = root.left
            else:
                top = stk[-1]
                if not top.right or top.right == pre:
                    res.append(top.val)
                    pre = top
                    stk.pop()
                else:
                    root = top.right
        return res

跟上Binary Tree Preorder Traversal.