题目描述:

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

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

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1
 \
  2
 /
3

return [1,2,3].

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

二叉树非递归前序遍历. 需要背下来的代码了…

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/**
 * 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> preorderTraversal(TreeNode *root) {
        vector<TreeNode*> path;
        vector<int> ret;
        if(root == NULL) return ret;
        TreeNode *node = root;

        while(node || !path.empty()){
            if(node){
                ret.push_back(node->val);
                path.push_back(node);
                node = node->left;
            }
            else{
                node = path.back();
                path.pop_back();
                node = node->right;
            }
        }

        return ret;
    }
};