[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。

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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.csdn.net/qq_14821023/article/details/50809788

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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;    }};