LeetCode之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?

//不断访问其左子树，然后输出其根，然后访问其接着的右子树，重复过程

#include <stack>
#include <vector>
/**
 * 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) {
        stack <TreeNode *> intStack;
		vector<int> intVector;
		//if (!root) return intVector;
		TreeNode *p = root;
		do{
			while (p){
				intStack.push(p);
				p = p->left;
			}
			if (!intStack.empty()){
			    TreeNode *tmp=intStack.top();
				intVector.push_back(tmp->val);
				intStack.pop();
				p = tmp->right;
			}
		} while (!intStack.empty() || p);
		return intVector;   
    }
};

LeetCode之Binary Tree Inorder Traversal,布布扣,bubuko.com

LeetCode之Binary Tree Inorder Traversal

原文：http://blog.csdn.net/smileteo/article/details/21643263