【Leetcode】145. 二叉树的后序遍历

题目描述

给定一个二叉树，返回它的 后序 遍历。

题解

递归法

执行用时：0 ms, 在所有 Java 提交中击败了100.00%的用户

内存消耗：36.8 MB, 在所有 Java 提交中击败了29.78%的用户

/**
 * 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 {
    List<Integer> res;

    public List<Integer> postorderTraversal(TreeNode root) {
        this.res = new ArrayList<>();
        postorder(root);
        return res;
    }

    public void postorder(TreeNode root) {
        if (root == null)
            return;
        postorder(root.left);
        postorder(root.right);
        res.add(root.val);
    }
}
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非递归写法：

执行用时：1 ms, 在所有 Java 提交中击败了29.35%的用户

内存消耗：36.7 MB, 在所有 Java 提交中击败了33.92%的用户

/**
 * 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 {
    List<Integer> res;

    public List<Integer> postorderTraversal(TreeNode root) {
        this.res = new ArrayList<>();
        if (root == null)
            return res;
        Stack<TreeNode> stack = new Stack<>();
        while (root != null || !stack.isEmpty()) {
            if (root != null) {
                stack.push(root);
                res.add(0, root.val);
                root = root.right;
            }
            else {
                root = stack.pop();
                root = root.left;
            }
        }
        return res;
    }
}
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