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94 Binary Tree Inorder Traversal


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


For example:
Given binary tree {1,#,2,3},
   1
    \
     2
    /
   3
return [1,3,2].


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

//方法一：堆栈
/**
 * Definition for binary tree
 * 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> ans;
        ans.clear();
        if(root==NULL)
            return ans;
        stack<TreeNode *> st;
        TreeNode *p = root;
        while( p!=NULL || !st.empty() )
        {
            while( p!=NULL )
            {
                st.push(p);
                p = p->left;
            }
            if(!st.empty())
            {
                p = st.top();
                st.pop();
                ans.push_back(p->val);
                p = p->right;
            }
        }
        return ans;
    }
};

//方法二：递归
class Solution {
vector<int> ans;
public:
    void inorderTraversalUtil(TreeNode *root)
    {
        if(root==NULL)
            return;
        inorderTraversalUtil(root->left);
        ans.push_back(root->val);
        inorderTraversalUtil(root->right);
    }
    
    vector<int> inorderTraversal(TreeNode *root) {
        ans.clear();
        if(root == NULL)
            return ans;
        inorderTraversalUtil(root);
        return ans;
    }
};

leetcode_94_Binary Tree Inorder Traversal

标签：c++   leetcode   tree   recursion   

原文地址：http://blog.csdn.net/keyyuanxin/article/details/44216847