Monday, June 16, 2014

Binary Tree Preorder Traversal@leetcode

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?
用stack做,res存结果,不要混淆,那么我们要的是res的顺序是preorder, 那么stack是一个缓冲,root, left, right, 那么stack是root, right, left, 用一个while循环来模拟.
/**
* 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<int> res;
if(root == NULL) return res;
stack<TreeNode*> stack;
stack.push(root);
while(!stack.empty()){
res.push_back(stack.top()->val);
TreeNode* tmp = stack.top();
stack.pop();
if(tmp->right) stack.push(tmp->right);
if(tmp->left) stack.push(tmp->left);
}
return res;
}
};

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