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?
/**
* 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> result;
TreeNode *p = root;
stack<TreeNode *>s;
while (!s.empty() || p != NULL) {
if (p != NULL) {
s.push(p);
p = p->left;
}
else {
p = s.top();
result.push_back(p->val);
s.pop();
p = p->right;
}
}
return result;
}
};

本文介绍了一种不使用递归实现二叉树中序遍历的方法。通过运用栈来辅助处理节点,避免了传统递归方法可能带来的栈溢出问题,特别适用于大规模数据处理场景。
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