145. Binary Tree Postorder Traversal

难度: Easy

题目描述

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]

Explanation:

Example 2:

Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]

Output: [4,6,7,5,2,9,8,3,1]

Explanation:

Example 3:

Input: root = []

Output: []

Example 4:

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?

解题思路

代码实现

解决方案

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;
 * }
 * }
 */
class Solution {
    // private void postOrder(TreeNode root, List<Integer> res) {
    // if (root == null) {
    // return;
    // }
    // postOrder(root.left, res);
    // postOrder(root.right, res);
    // res.add(root.val);
    // }

    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new LinkedList<>();
        TreeNode cur = root;
        LinkedList<TreeNode> stk = new LinkedList<>();
        while (cur != null || !stk.isEmpty()) {

            if (cur != null) {
                res.add(cur.val);
                if (cur.left != null) {
                    stk.push(cur.left);
                }
                cur = cur.right;
            } else {
                cur = stk.pop();
            }

            // while (cur != null) {
            // res.add(cur.val);
            // if (cur.left != null) {
            // stk.push(cur.left);
            // }

            // cur = cur.right;

            // }

            // if (!stk.isEmpty()) {
            // cur = stk.pop();
            // }

        }

        Collections.reverse(res);
        return res;

    }
}