LeetCode 94. Binary Tree Inorder Traversal（二叉树的中序遍历）

最新推荐文章于 2022-10-16 15:54:21 发布

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本文介绍了一种解决二叉树中序遍历的方法，包括递归和迭代两种方式。递归方式直观易懂，而迭代方式则利用栈结构避免了递归调用。此外，还提供了一种通过节点状态判断进行优化的迭代方法。

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原题网址：https://leetcode.com/problems/binary-tree-inorder-traversal/

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

For example:
Given binary tree {1,#,2,3},

return [1,3,2].

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

方法一：递归。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private List<Integer> traversal = new ArrayList<>();
    private void traverse(TreeNode root) {
        if (root == null) return;
        if (root.left != null) traverse(root.left);
        traversal.add(root.val);
        if (root.right != null) traverse(root.right);
    }
    public List<Integer> inorderTraversal(TreeNode root) {
        traverse(root);
        return traversal;
    }
}

方法二：将方法一的递归之间修改为用栈实现。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> traversal = new ArrayList<>();
        Stack<TreeNode> nodes = new Stack<>();
        Stack<Integer> states = new Stack<>();
        nodes.push(root);
        states.push(1);
        while (!nodes.isEmpty()) {
            TreeNode node = nodes.pop();
            int state = states.pop();
            if (node == null) continue;
            if (state == 1) {
                nodes.push(node);
                states.push(2);
                nodes.push(node.left);
                states.push(1);
                continue;
            } else if (state == 2) {
                traversal.add(node.val);
                nodes.push(node);
                states.push(3);
                nodes.push(node.right);
                states.push(1);
                continue;
            } else {
                // nothing to do
            }
        }
        return traversal;
    }
}

方法三：通过prev来区分当前是正在向下遍历，还是已经遍历完成，以免重复压栈。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> traversal = new ArrayList<>();
        if (root == null) return traversal;
        Stack<TreeNode> nodes = new Stack<>();
        TreeNode prev = null;
        nodes.push(root);
        while (!nodes.isEmpty()) {
            TreeNode node = nodes.pop();
            if (prev == null || node == prev.left || node == prev.right) {
                if (node.left != null && node.right != null) {
                    nodes.push(node.right);
                    nodes.push(node);
                    nodes.push(node.left);
                } else if (node.left != null) {
                    nodes.push(node);
                    nodes.push(node.left);
                } else if (node.right != null) {
                    traversal.add(node.val);
                    nodes.push(node.right);
                } else { 
                    traversal.add(node.val);
                }
            } else {
                traversal.add(node.val);
            }
            prev = node;
        }
        return traversal;
    }
}

简化：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> results = new ArrayList<>();
        if (root == null) return results;
        ArrayDeque<TreeNode> stack = new ArrayDeque<>();
        stack.push(root);
        TreeNode prev = null;
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            if (prev == null || prev.left == node || prev.right == node) {
                if (node.right != null) stack.push(node.right);
                if (node.left != null) {
                    stack.push(node);
                    stack.push(node.left);
                } else {
                    results.add(node.val);
                }
            } else {
                results.add(node.val);
            }
            prev = node;
        }
        return results;
    }
}