LeetCode——Binary Tree Inorder Traversal-优快云博客

本文链接：https://blog.youkuaiyun.com/pickless/article/details/11751175

本文详细介绍了如何通过使用堆栈替代递归来实现二叉树的中序遍历，提供了不同于传统的递归解决方案的迭代方法，并解释了堆栈在该过程中的应用。

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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?

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}".

» Solve this problem

题目的输入并不是一个Serialized Tree，还是普通的Tree。

该题的关键在于不用递归，那么自然会想到用堆栈来替代递归过程。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> ans;
        if (root == NULL) {
            return ans;
        }
        stack<TreeNode*> stack;
        stack.push(root);
        while (!stack.empty()) {
            TreeNode* node = stack.top();
            if (node->left == NULL && node->right == NULL) {
                stack.pop();
                ans.push_back(node->val);
            }
            else {
                stack.pop();
                if (node->right != NULL) {
                    stack.push(node->right);
                }
                stack.push(node);
                if (node->left != NULL) {
                    stack.push(node->left);
                }
                node->left = node->right = NULL;
            }
        }
        return ans;
    }
};