二叉树的后序遍历 (Binary Tree Postorder Traversal)

 

思路:递归

// @Title: 二叉树的后序遍历 (Binary Tree Postorder Traversal)
// @Author: qisiii
// @Date: 2024-09-14 21:05:17
// @Runtime: 0 ms
// @Memory: 40.7 MB
// @comment: 递归
// @flag: GREEN
/**
 * 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 {
    public List<Integer> postorderTraversal(TreeNode root) {
    List<Integer> result=new ArrayList<>();
        sort(result,root);
        return result;
    }
    private void sort(List<Integer> result,TreeNode node){
        if(node!=null){
            sort(result,node.left);
            sort(result,node.right);
            result.add(node.val);
        }
    }
}

+++ title = “二叉树的后序遍历 (Binary Tree Postorder Traversal)” draft = false +++

思路:迭代

// @Title: 二叉树的后序遍历 (Binary Tree Postorder Traversal)
// @Author: qisiii
// @Date: 2024-09-14 22:35:50
// @Runtime: 1 ms
// @Memory: 40.8 MB
// @comment: 迭代
// @flag: GREEN
/**
 * 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 {
    public List<Integer> postorderTraversal(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        List<Integer> result = new ArrayList<>();
        HashSet<TreeNode> parent = new HashSet<>();
        while (root != null || !stack.isEmpty()) {
            if (root == null && parent.contains(stack.peek())) {
                //当左子树和右子树都处理完了
                result.add(stack.pop().val);
            } else if (root == null) {
                parent.add(stack.peek());
                //然后处理右子树
                root = stack.peek().right;
            } else {
                //先将左边的节点都放到队列
                stack.push(root);
                root = root.left;
            }
        }
        return result;
    }
}
Licensed under CC BY-NC-SA 4.0
最后更新于 2024-10-18