LeetCode-145. Binary Tree Postorder TraversalLevel up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.https://leetcode.com/problems/binary-tree-postorder-traversal/

Given the root of a binary tree, return the postorder traversal of its nodes' values.

Example 1:

Input: root = [1,null,2,3]
Output: [3,2,1]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

Constraints:

• The number of the nodes in the tree is in the range [0, 100].
• -100 <= Node.val <= 100

Follow up: Recursive solution is trivial, could you do it iteratively?

【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) {}
 * };
 */

1. 双栈

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> postorder;
        if (root == nullptr) {return postorder;}
        stack st1, st2;
        st1.push(root);
        while (!st1.empty()){
            root = st1.top();
            st1.pop();
            st2.push(root);
            if (root->left != nullptr) {st1.push(root->left);}
            if (root->right != nullptr) {st1.push(root->right);}
        }
        while (!st2.empty()){
            postorder.push_back(st2.top()->val);
            st2.pop();
        }
        return postorder;
    }
};

2. 单栈反序 

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        if(!root) return res;
        stack st;
        st.push(root);
        while(!st.empty()) {
            TreeNode* temp = st.top(); st.pop();
            if(temp->left) st.push(temp->left);
            if(temp->right) st.push(temp->right);
            res.push_back(temp->val);
        }
        reverse(res.begin(),res.end());
        return res;
    }
};

【Java】

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */

1. 双栈

class Solution {
    public List postorderTraversal(TreeNode root) {
        List postorder = new ArrayList<>();
        if (root == null) {return postorder;}
        Stack st1 = new Stack<>();
        Stack st2 = new Stack<>();
        st1.push(root);
        while (!st1.isEmpty()){
            root = st1.pop();
            st2.push(root);
            if (root.left != null) {st1.push(root.left);}
            if (root.right != null) {st1.push(root.right);}
        }
        while (!st2.isEmpty()){
            postorder.add(st2.pop().val);
        }
        return postorder;
    }
}