leetcode解题: Binary Tree Inorder Traversal

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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?

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解法1： Recursion

用递归的in-order很简单，按字面理解就是先左再自己再右。
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/**
 * 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> res;
        helper(root, res);
        return res;
    }
    void helper(TreeNode* root, vector<int>& res) {
        if (!root) {
            return;
        }
        helper(root->left, res);
        res.push_back(root->val);
        helper(root->right, res);
    }
};

Java

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解法2： Iterative

主要考虑的就是用一个stack来存储还没有访问的节点， 难点在于确定什么时候push和pop。
那么在寻找最小值的时候，一路上碰到的所有node都push
如果碰到了空节点，则证明现在的这条路已经到底了，需要往回寻找，这个时候就可以pop了，记录节点的数值。并且把将要探寻的指针放到right上继续。
C++

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/**
 * 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> res;
        if (!root) {
            return res;
        }
        stack<TreeNode*> store;
        TreeNode* cur = root;
        while (cur || !store.empty()) {
            if (cur) {
                store.push(cur);
                cur = cur->left;
            } else {
                TreeNode* temp = store.top();
                res.push_back(temp->val);
                store.pop();
                cur = temp->right;
            }
        }
        return res;
    }
};

Java

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/**
 * 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<>();
        while (root!= null || !stack.isEmpty()) {
            if (root != null) {
                stack.push(root);
                root = root.left;
            } else {
                TreeNode temp = stack.pop();
                res.add(temp.val);
                root = temp.right;
            }
        }
        return res;
    }
}