【leetcode】94. Binary Tree Inorder Traversal 二叉树的中序遍历

1. 题目

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

For example:
Given binary tree [1,null,2,3],
1

\
 2
/

3
return [1,3,2].

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

2. 思路

递归相对简单, 递归左子树，压入根节点，递归右子树，结束。

模拟遍历的过程。迭代的方式采用栈来记录遍历的路径。初始将root压入。
然后每次从栈顶开始，持续将栈顶的左子树非空的左子树压栈。如果为空，则表示达到了当前的最左侧，输出当前节点。让后持续向上并输出，找到第一个右子树非空的情况进行压栈。
栈空的时候表示全部遍历完成。

3. 代码-递归

/**
 * Definition for a binary tree node.
 * 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) {
        vector<int> ret;
        if (root == NULL) { return ret; }
        //if (root->left == NULL && root->right == NULL) { ret.push_back(root->val); return ret; }
        vector<int> left = inorderTraversal(root->left);
        vector<int> right = inorderTraversal(root->right);
        left.push_back(root->val);
        for (auto it = right.begin(); it != right.end(); it++) {
            left.push_back(*it);
        }
        return left;
    }
};

4. 代码-迭代

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    TreeNode* _stack[1024];
    int _top;
public:
    vector<int> inorderTraversal(TreeNode* root) {
        _top = 0;
        vector<int> ret;
        if (root == NULL) { return ret; }
        _stack[_top] = root;
        while (_top >= 0) {
            if (_stack[_top]->left != NULL) {
                _stack[_top+1] = _stack[_top]->left;
                _top++;
            } else {
                ret.push_back(_stack[_top]->val);
                if (_stack[_top]->right != NULL) {
                    _stack[_top] = _stack[_top]->right;
                } else {
                    _top--;
                    while (_stack[_top]->right == NULL && _top >= 0) {
                        ret.push_back(_stack[_top]->val);
                        _top--;
                    }
                    if (_top >= 0) {
                        ret.push_back(_stack[_top]->val);
                        _stack[_top] = _stack[_top]->right;
                    }
                }
            }
        }
        return ret;
    }
};