LeetCode94. 二叉树的中序遍历

递归代码展示:

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

非递归(栈实现)代码展示:

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

代码解释:一直顺着根节点的左结点找下去,直到某个节点的左结点为空,把这个结点的值压栈,然后访问这个节点的右节点。再以同样的方式顺着这个节点的左结点找下去……


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转载自blog.csdn.net/jaster_wisdom/article/details/80918635