1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51

/*
 * @lc app=leetcode id=145 lang=java
 *
 * [145] Binary Tree Postorder Traversal
 *
 * https://leetcode.com/problems/binary-tree-postorder-traversal/description/
 *
 * algorithms
 * Easy (70.88%)
 * Likes:    6835
 * Dislikes: 189
 * Total Accepted:    1.2M
 * Total Submissions: 1.7M
 * Testcase Example:  '[1,null,2,3]'
 *
 * 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?
 */

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72

/*
 * 思路1：递归实现后序遍历
*/
// class Solution1 {
//     List<Integer> result = new ArrayList<>();
//     public List<Integer> postorderTraversal(TreeNode root) {
//         if(null == root){
//             return result;
//         }
//         postorderTraversal(root.left);
//         postorderTraversal(root.right);
//         result.add(root.val);
//         return result;   
//     }
// }

/*
 * 思路2：stack递归实现后序遍历，使用一个额外的栈反转节点
*/
class Solution2 {
    List<Integer> result = new ArrayList<>();

    public List<Integer> postorderTraversal(TreeNode root) {
        if(root == null){
            return result;
        }
        Stack<TreeNode> stack1 = new Stack<>();
        Stack<TreeNode> stack2 = new Stack<>();
        stack1.push(root);
        while (!stack1.isEmpty()) {
            TreeNode node = stack1.pop();
            stack2.push(node);
            if(node.left != null){
                stack1.push(node.left);
            }
            if(node.right != null){
                stack1.push(node.right);
            }
        }
        while (!stack2.isEmpty()) {
            result.add(stack2.pop().val);
        }
        return result;
    }
}


/*
 * 思路3：stack递归实现后序遍历，借鉴谦虚遍历，稍微调整下入栈顺序即可
*/
class Solution {
    List<Integer> result = new ArrayList<>();
    public List<Integer> postorderTraversal(TreeNode root) {
        if(root == null){
            return result;
        }
        Stack<Integer> stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            result.add(node.val);
            if(root.left != null){
                stack.push(root.left);
            } 
            if(root.right != null){
                stack.push(root.left);
            }
        }
        result.reversed();
        return result;
    }
}