94 二叉树中序遍历

 

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].

 

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        res = []
        if root == None:
            return res
        stack = []
        cur = root
        while cur or stack:
            while cur:
                stack.append(cur)
                cur = cur.left
            t = stack.pop()
            res.append(t.val)
            cur = t.right
        return res

 

 

 

非递归：

 

 

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