[LeetCode Hot 100] LeetCode145. 二叉树的后序遍历

题目描述

思路

• 递归：额外写一个函数void postOrder(TreeNode node, List res)
• 迭代：
  • 前序遍历：根---左---右
  • 将前序遍历改造成：根---右---左
  • 然后反转根右左为：左---右---根，即为后序遍历
  • 优化一下：

while (!stack.isEmpty()) {
	TreeNode node = stack.pop();
	res.addFirst(node.val);
	if (node.left != null) stack.push(node.left);
	if (node.right != null) stack.push(node.right);
}

方法一：递归

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

方法二：迭代

/**
 * 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> res = new ArrayList<>();
        Deque<TreeNode> stack = new ArrayDeque<>();
        if (root == null) return res;
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            res.add(node.val);
            if (node.left != null) stack.push(node.left);
            if (node.right != null) stack.push(node.right);
        }
        Collections.reverse(res);
        return res;
    }
}

优化一下：

/**
 * 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> res = new ArrayList<>();
        Deque<TreeNode> stack = new ArrayDeque<>();
        if (root == null) return res;
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            res.addFirst(node.val);
            if (node.left != null) stack.push(node.left);
            if (node.right != null) stack.push(node.right);
        }
        //Collections.reverse(res);
        return res;
    }
}