标签：
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?
中序遍历，左-根-右
/**
* 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 { //中序遍历 左-根-右
public:
vector<int> res;
vector<int> inorderTraversal(TreeNode* root) {
if(root!=NULL)
{
inorderTraversal(root->left);
res.push_back(root->val);
inorderTraversal(root->right);
}
return res;
}
};/**
* 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 { //中序遍历 左-根-右
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> res;
stack<TreeNode*> stack;
TreeNode* pNode=root;
while(pNode||!stack.empty())
{
if(pNode!=NULL) //节点不为空，加入栈中，并访问节点左子树
{
stack.push(pNode);
pNode=pNode->left;
}
else
{
pNode=stack.top(); //节点为空，从栈中弹出一个节点，访问这个节点
stack.pop();
res.push_back(pNode->val);
pNode=pNode->right; //访问节点右子树
}
}
return res;
}
};leetcode No94. Binary Tree Inorder Traversal
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原文地址：http://blog.csdn.net/u011391629/article/details/52201639
