[LeetCode] 144. Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.

Example:

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

Output: [1,2,3]

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

二叉树的先序遍历。

树的几种遍历，网上有一个不错的例子，一个树给了多种遍历的结果

Depth First Traversals:
(a) Inorder (Left, Root, Right) : 4 2 5 1 3
(b) Preorder (Root, Left, Right) : 1 2 4 5 3
(c) Postorder (Left, Right, Root) : 4 5 2 3 1

Breadth First or Level Order Traversal : 1 2 3 4 5

对于这题，我在网上又找了一个更复杂的树及其遍历结果，如下图（reference）

二叉树的遍历有多种方式，此题是要求用先序遍历。我自己总结的先序遍历的特点是根 - 左 - 右。即如果只有一个根节点和两个子节点的情形下（比如[1, 2, 3]），输出结果应该也是[1, 2, 3]。这题迭代和递归都需要掌握。遍历的时候会用到栈，因为栈是先进后出，所以放进栈的时候记得要逆向，先放右孩子再放左孩子。两种做法的时间复杂度是O(n)，空间复杂度是O(h)。h是树的高度。

迭代

JavaScript实现

 1 /**
 2  * @param {TreeNode} root
 3  * @return {number[]}
 4  */
 5 var preorderTraversal = function(root) {
 6     // corner case
 7     if (!root) return [];
 8 
 9     // normal case
10     let result = [];
11     // let stack = [root];
12     let stack = [];
13     stack.push(root);
14     while (stack.length) {
15         var node = stack.pop();
16         result.push(node.val);
17         if (node.right) stack.push(node.right);
18         if (node.left) stack.push(node.left);
19     }
20     return result;
21 };

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public List<Integer> preorderTraversal(TreeNode root) {
12         List<Integer> res = new ArrayList<>();
13         if (root == null) {
14             return res;
15         }
16         Stack<TreeNode> stack = new Stack<>();
17         stack.push(root);
18         while (!stack.isEmpty()) {
19             TreeNode cur = stack.pop();
20             res.add(cur.val);
21             if (cur.right != null) {
22                 stack.push(cur.right);
23             }
24             if (cur.left != null) {
25                 stack.push(cur.left);
26             }
27         }
28         return res;
29     }
30 }

递归

JavaScript实现

 1 /**
 2  * @param {TreeNode} root
 3  * @return {number[]}
 4  */
 5 var preorderTraversal = function(root) {
 6     let res = [];
 7     if (root === null) {
 8         return res;
 9     }
10     helper(res, root);
11     return res;
12 };
13 
14 var helper = function(res, root) {
15     if (root === null) return;
16     res.push(root.val);
17     helper(res, root.left);
18     helper(res, root.right);
19 };

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public List<Integer> preorderTraversal(TreeNode root) {
12         List<Integer> res = new ArrayList<>();
13         if (root == null) return res;
14         helper(res, root);
15         return res;
16     }
17 
18     private void helper(List<Integer> res, TreeNode root) {
19         if (root == null) return;
20         res.add(root.val);
21         helper(res, root.left);
22         helper(res, root.right);
23     }
24 }

树的遍历

94. Binary Tree Inorder Traversal

144. Binary Tree Preorder Traversal

145. Binary Tree Postorder Traversal

LeetCode 题目总结