二叉樹的中序遍歷

leetcode原題：94. 二叉樹的中序遍歷

題目描述

給定一個二叉樹的根節點root，返回它的中序遍歷。

示例 1：

輸入：root = [1,null,2,3]
輸出：[1,3,2]

示例 2：

輸入：root = []
輸出：[]

示例 3：

輸入：root = [1]
輸出：[1]

示例 4：

輸入：root = [1,2]
輸出：[2,1]

示例 5：

輸入：root = [1,null,2]
輸出：[1,2]

提示：

• 樹中節點數目在范圍[0, 100]內
• -100 <= Node.val <= 100

進階: 遞歸算法很簡單，你可以通過迭代算法完成嗎？

分析

遞歸思路：

先遞歸訪問左孩子，然后記錄該節點的值，再遞歸訪問右孩子。

迭代思路：

中序遍列，始終牢記【左，根，右】的順序，優先左孩子，然后根節點，最后右孩子，遍歷左右孩子時同樣遵循【左，根，右】的順序。

1. 只要有左孩子，就需要一直往左遍歷，所以需要用棧來記錄遇到的節點，直到左孩子為空為止；
2. 彈出棧頂元素，並輸出該節點的值；
3. 當第2步中的棧頂節點的右孩子為空，則重復步驟2和3，否則重復步驟1,2,3。

實現

Java

遞歸版本：

/**
 * 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<>();
        visit(res, root);
        return res;
    }

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

迭代版本：

/**
 * 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;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        LinkedList<TreeNode> stack = new LinkedList<>();

        while (!queue.isEmpty() || !stack.isEmpty()) {

            if (!queue.isEmpty()) {
                TreeNode cur = queue.poll();
                stack.push(cur);

                while (cur.left != null) {
                    cur = cur.left;
                    stack.push(cur);
                }
            }

            if (!stack.isEmpty()) {
                TreeNode cur = stack.pop();
                res.add(cur.val);

                if (cur.right != null) {
                    queue.offer(cur.right);
                }
            }
        }

        return res;
    }
}

迭代過程中，你可以發現上面代碼的那個queue中其實始終最多只會有一個元素，所以可以再簡單優化一下：

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

        while (cur != null || !stack.isEmpty()) {
            if (cur != null) {
                stack.push(cur);
                cur = cur.left;
            } else {
                cur = stack.pop();
                res.add(cur.val);
                cur = cur.right;
            }
        }
        return res;
    }
}

Python3

遞歸版本：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        res = []

        def visit(root: TreeNode):
            if root:
                visit(root.left)
                res.append(root.val)
                visit(root.right)

        visit(root)
        return res

迭代版本：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        res = []

        cur = root
        stack = []

        while cur or stack:
            if cur:
                stack.append(cur)
                cur = cur.left
            else:
                cur = stack.pop()
                res.append(cur.val)
                cur = cur.right

        return res