leetcode94 - Binary Tree Inorder Traversal - medium

Given a binary tree, return the inorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,3,2]

Follow up: Recursive solution is trivial, could you do it iteratively?

 

Inorder: root in the middle

先一路traverse到最左边的leaf node，一路上把经过的node都push进stack。最先pop出来的就是起点，也相当于一个夹在中间的root（它的left child是null），所以去到它的right child，再如此循环。因为不涉及修改node，直接拿root来traverse就行，stack里都是相对来说的left node。

 

实现：

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        
        vector<int> res;
        if (!root) return res;
        
        stack<TreeNode*> st;
        while (!st.empty() || root){
            while (root){
                st.push(root);
                root = root->left;
            }
            root = st.top();
            st.pop();
            res.push_back(root->val);
            root = root->right;
        }
        
        return res;
        
    }
};