[LeetCode] 94. Binary Tree Inorder Traversal

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

Example 1:

Input: root = [1,null,2,3]
Output: [1,3,2]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

Example 4:

Input: root = [1,2]
Output: [2,1]

Example 5:

Input: root = [1,null,2]
Output: [1,2]

Constraints:

• The number of nodes in the tree is in the range [0, 100].
• -100 <= Node.val <= 100

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

二叉树的中序遍历。

中序遍历我记为 左 - 根 - 右。

还是用GeeksforGeeks的例子来描述如何做中序遍历吧。

Inorder (Left, Root, Right) : 4 2 5 1 3

树的遍历大部分都是可以给出迭代和递归两种做法的，此题我也给出两种做法。两种做法的时间和空间复杂度一样，时间都是O(n)，空间都是O(h)。

迭代

（配合例子看）迭代的做法是需要将所有的左孩子先放进栈，这样弹出的时候（15行），最小的左孩子4就被存入结果集了。此时cur指针在stack顶端，2的位置，所以此时才会有17行去看2是否有右孩子。有的话，下一次while循环的一开头，这个右孩子5就会被再次放进栈。

JavaScript实现

/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var inorderTraversal = function(root) {
    let res = [];
    if (root === null) return res;
    let stack = [];
    let cur = root;
    while (cur !== null || stack.length) {
        while (cur !== null) {
            stack.push(cur);
            cur = cur.left;
        }
        cur = stack.pop();
        res.push(cur.val);
        cur = cur.right;
    }
    return res;
};

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

递归

递归的做法很直观，既然是中序遍历，所以根节点需要放在左孩子和右孩子的helper函数中间。

JavaScript实现

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

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

Java实现

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

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

关于为什么很多树的题目的空间复杂度是O(h) - 树的高度，我这里给一个discussion里面看到的截图。我看到这个截图我彻底懂了为什么是跟树的高度有关了。

树的遍历

94. Binary Tree Inorder Traversal

144. Binary Tree Preorder Traversal

145. Binary Tree Postorder Traversal

LeetCode 题目总结