【LeetCode】144. 二叉树的前序遍历

144. 二叉树的前序遍历

知识点：二叉树；递归；Morris遍历

题目描述

给你二叉树的根节点 root ，返回它节点值的 前序 遍历。

示例

输入：root = [1,null,2,3]
输出：[1,2,3]

输入：root = []
输出：[]

输入：root = [1]
输出：[1]

输入：root = [1,2]
输出：[1,2]

输入：root = [1,null,2]
输出：[1,2]

---

解法一：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 {
    List<Integer> list = new ArrayList<>();
    public List<Integer> preorderTraversal(TreeNode root) {
        if(root == null) return list;
        list.add(root.val);
        preorderTraversal(root.left);
        preorderTraversal(root.right);
        return list;
    }
}

时间复杂度；0(N),每个节点恰好被遍历一次；

空间复杂度；O(N),递归过程中栈的开销；

解法二:迭代法

入栈一定是先右后左，这样出来才能使先左后右；

压入根节点；

1.弹出就打印；

2.如有右孩子，压入右；

3.如有左孩子，压入左；

/**
 * 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> preorderTraversal(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        List<Integer> list = new ArrayList<>();
        if(root != null) stack.push(root);
        while(!stack.isEmpty()){
            TreeNode top = stack.pop();
            list.add(top.val);
            if(top.right != null) stack.push(top.right);
            if(top.left != null) stack.push(top.left);
        }
        return list;
    }
}

解法三：Morris遍历

构建从下到上的连接，一条路能够走遍所有节点；

/**
 * 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> list = new ArrayList<>();
    public List<Integer> preorderTraversal(TreeNode root) {
        if(root == null) return list;
        TreeNode cur = root;
        TreeNode mostRightNode = null;
        while(cur != null){
            mostRightNode = cur.left;
            if(mostRightNode != null){
                //有左子树；
                while(mostRightNode.right != null && mostRightNode.right != cur){
                    mostRightNode = mostRightNode.right; //找到左子树的最右节点；
                }
                if(mostRightNode.right == null){
                    mostRightNode.right = cur; //构建向上的连接；
                    list.add(cur.val);
                    cur = cur.left;
                    continue;
                }else{
                    //第二次到节点，断开连接
                    mostRightNode.right = null;
                    cur = cur.right;
                }
            }else{
                list.add(cur.val);
                cur = cur.right;
            }
        }
        return list;
    }
}

体会

二叉树的遍历是二叉树的最基础的，不仅要掌握递归写法，迭代法和morris遍历也需要掌握，更多详细可参考这篇总结二叉树

相关题目

94. 二叉树的中序遍历

145. 二叉树的后序遍历

相关链接

二叉树