[LeetCode-11] Binary Tree Inorder Traversal-优快云博客

本文链接：https://blog.youkuaiyun.com/sbitswc/article/details/26022051

本文介绍了一种实现二叉树中序遍历的方法，包括递归和迭代两种方式。递归方法简单直观，而迭代方法则使用栈来辅助完成。通过这两种方法可以有效地获取二叉树节点值的中序遍历结果。

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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?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

For recursion version, it's very easy to write.

But for iterative version, we need a stack to help.

java recursion

public ArrayList<Integer> result;
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        result = new ArrayList<Integer>();
        inorderTrRe(root);
		return result;
    }
    public void inorderTrRe(TreeNode root){
		if(root!=null){
        	inorderTrRe(root.left);
        	result.add(root.val);
        	inorderTrRe(root.right);
        }
	}

java iterative

public class Solution {
    public ArrayList<Integer> result;
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        result = new ArrayList<Integer>();
        if(root== null) return result;
        inorderTraversal1(root);
		return result;
    }
    public void inorderTraversal1(TreeNode root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
		Stack<TreeNode> stack = new Stack<TreeNode>();
		TreeNode node = root;
		while(stack.size()>0 || node!=null){
			while(node != null){
				stack.push(node);
				node = node.left;
			}
			node = stack.pop();
			result.add(node.val);
			node = node.right;
		}
    }
}

c++ recursion

void inorder(TreeNode *root,vector<int> &path){
        if(root!=NULL){
            inorder(root->left,path);
            path.push_back(root->val);
            inorder(root->right,path);
        }
    }
    vector<int> inorderTraversal(TreeNode *root) {
    vector<int> path;
    if(root == NULL) return path;
    inorder(root, path);
    return path;
    }

c++ iterative

vector<int> inorderTraversal(TreeNode *root) {
        vector<int> path;
    stack<TreeNode*> stk;
    path.clear();
    if(root == NULL) return path;
    TreeNode *p = root;
    while(p != NULL || !stk.empty()){
        while(p!=NULL){
            stk.push(p);
            p = p->left;
        }
            p = stk.top();
            stk.pop();
            path.push_back(p->val);
            p = p->right;
        
    }
    return path;
    }