leetcode: Binary Tree Preorder Traversal

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

For example:
Given binary tree {1,#,2,3},

   1
         2
    /
   3

return [1,2,3].

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

本题为二叉树的先序遍历，也就是先根遍历，但是提示不能用递归，所以就是二叉树先序遍历的非递归实现。现在为了在面试中能够灵活的掌握非递归遍历，三种遍历方法都总结一下，方面以后的面试中复习使用。

先序遍历：

  递归：

void traverse(Tree *root)
{
       if(root==NULL) return;
       visit(root->data);
       traverse(Tree->left);
       traverse(Tree->right);
}

void traverse(Tree *root)
{
       stack.push(root);
       while(!stack.empty())
     {
              temp=stack.top();
              stack.pop();
              visit(temp->data);
              stack.push(temp->right);//先将右儿子入栈，因为比左儿子更晚访问到
              stack.push(temp->left);
     }
}

递归：

 void   traverse(Tree *root){

                if(root==NULL)    return;

                traverse(Tree->left);

                visit(root->data);

                traverse(Tree->right);

        }                       

递归：

 void   traverse(Tree *root){

                if(root==NULL)    return;

                traverse(Tree->left);


                traverse(Tree->right);

                 visit(root->data);

        }                       

/**
 * 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> preorderTraversal(TreeNode* root) {
        TreeNode* temp;
        stack<TreeNode *> st;
        st.push(root);
        while(!st.empty())
        {
            temp=st.top();
            st.pop();
            
            visit(temp->val);
            st.push(temp->right);
            st.push(temp->left);
        }
    }
};

leetcode: Binary Tree Preorder Traversal

原文：http://blog.csdn.net/flyljg/article/details/47426837