[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

/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var preorderTraversal = function(root) {
    // corner case
    if (!root) return [];

    // normal case
    let result = [];
    // let stack = [root];
    let stack = [];
    stack.push(root);
    while (stack.length) {
        var node = stack.pop();
        result.push(node.val);
        if (node.right) stack.push(node.right);
        if (node.left) stack.push(node.left);
    }
    return result;
};

Java实现

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

递归

JavaScript实现

/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var preorderTraversal = function(root) {
    let res = [];
    if (root === null) {
        return res;
    }
    helper(res, root);
    return res;
};

var helper = function(res, root) {
    if (root === null) return;
    res.push(root.val);
    helper(res, root.left);
    helper(res, root.right);
};

Java实现

/**
 * 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> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) return res;
        helper(res, root);
        return res;
    }

    private void helper(List<Integer> res, TreeNode root) {
        if (root == null) return;
        res.add(root.val);
        helper(res, root.left);
        helper(res, root.right);
    }
}

树的遍历

94. Binary Tree Inorder Traversal

144. Binary Tree Preorder Traversal

145. Binary Tree Postorder Traversal

LeetCode 题目总结