[LeetCode]94 二叉树中序遍历

Binary Tree Inorder Traversal(二叉树中序遍历)

【难度：Medium】
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?
使用迭代而不是递归的方法来中序遍历二叉树。

---

解题思路

整体思路与LeetCode 144题迭代方法先序遍历二叉树类似，中间细节稍微调整即可。
http://blog.youkuaiyun.com/qq_14821023/article/details/50809788

---

c++代码如下：

//迭代方法
/**
 * 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> v;

        stack<TreeNode*> s;
        TreeNode* cur = root;

        while (cur || !s.empty()) {
            if (cur) {
                //当前节点压栈
                s.push(cur);
                //先走左边
                cur = cur->left;
            } else {
                cur = s.top()->right;
                //左边走完后记录根节点的值
                v.push_back(s.top()->val);
                s.pop();
            }
        }

        return v;
    }
};

//递归方法
/**
 * 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> v;
        if (!root)
            return v;
        vector<int> tmp = inorderTraversal(root->left);
        if (!tmp.empty()) {
            for (int i = 0; i < tmp.size(); i++) {
                v.push_back(tmp[i]);
            }
        }
        v.push_back(root->val);

        vector<int> tmp2 = inorderTraversal(root->right);
        if (!tmp2.empty()) {
            for (int i = 0; i < tmp2.size(); i++) {
                v.push_back(tmp2[i]);
            }
        }
        return v;
    }
};