145. Binary Tree Postorder Traversal (H)

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

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3
Output: [3,2,1]

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

题意

求树的后序遍历。

思路

递归或迭代。

后序遍历的迭代方法和前序、中序的区别在于当前结点必须在左右子树全访问完后才能访问，这就导致了在利用栈进行回溯时不能随意出栈(只有结点被访问后才能出栈，而在回溯时当前结点还没有被访问)，因此需要借助一个辅助栈来进行回溯，用另一个栈来记录后序遍历的所有结点。

代码实现 - 递归

/**
 * 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> ans = new ArrayList<>();
        postOrder(root, ans);
        return ans;
    }
    private void postOrder(TreeNode x, List<Integer> ans) {
        if (x == null) {
            return;
        }
        postOrder(x.left, ans);
        postOrder(x.right, ans);
        ans.add(x.val);
    }
}

代码实现 - 迭代

/**
 * 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> ans = new ArrayList<>();
        Deque<TreeNode> temp = new ArrayDeque<>();        // 辅助栈，用来处理回溯
        Deque<TreeNode> stack = new ArrayDeque<>();        // 记录后序遍历的顺序
        TreeNode cur = root;
        while (cur != null || !temp.isEmpty()) {
            if (cur != null) {
                temp.push(cur);
                stack.push(cur);
                cur = cur.right;
            } else {
                cur = temp.pop().left;
            }
        }
        while (!stack.isEmpty()) {
            ans.add(stack.pop().val);
        }
        return ans;
    }
}