详细分析可参考LeetCode上的一篇博文。具体程序如下: 1vector<int> inorderTraversal(TreeNode*root) {2vector<int>rVec;3stack<TreeNode *>st;4TreeNode *tree =root;5while(tree || !st.empty())6{7if(tree)8{9st.push(tree);10tree = tree->left;11}12else13{14tree =st.top();15rVec.push...
*/// 由于主函数的形式已经符合分治法需要的形式(具有合适的返回值),直接使用主函数做为递归函数vector<int>preorderTraversal(TreeNode * root){//递归三要素之定义// write your code herevector<int> result;if(root == nullptr) {returnresult;// 递归三要素之出口}// 获取左子树的遍历结果vector<int> ...
*/voidtraversal(structTreeNode*root,int*index,int*res){if(!root)return;res[(*index)++]=root->val;traversal(root->left,index,res);traversal(root->right,index,res);}int*preorderTraversal(structTreeNode*root,int*returnSize){int*res=malloc(sizeof(int)*110);intindex=0;traversal(root,&index,...
1/**2* Definition for a binary tree node.3* struct TreeNode {4* int val;5* TreeNode *left;6* TreeNode *right;7* TreeNode(int x) : val(x), left(NULL), right(NULL) {}8* };9*/10classSolution {11public:12vector<int> preorderTraversal(TreeNode*root) {13vector<int>rVec;14pr...
LeetCode Binary Tree Preorder Traversal 先根遍历,题意:给一棵树,求其先根遍历的结果。思路:(1)深搜法:1/**2*Definitionforabinarytreenode.3*structTreeNode{4*intval;5*TreeNode*left;6*...
1#LeetCode 144. Binary Tree Preorder Traversal Given a binary tree, return the preorder traversal of its nodes' values. Note: Recursive solution is trivial, could you do it iteratively? Preorder: root, left, right Recursive version:/** ...
上边的两种解法,空间复杂度都是O(n),利用 Morris Traversal 可以使得空间复杂度变为O(1)。 它的主要思想就是利用叶子节点的左右子树是null,所以我们可以利用这个空间去存我们需要的节点,详细的可以参考94 题中序遍历。 publicList<Integer>preorderTraversal(TreeNoderoot){List<Integer>list=newArrayList<>();Tree...
同Construct Binary Tree from Inorder and Postorder Traversal(Python版) # Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def buildTree(self, pre...
144. Binary Tree Preorder Traversal Given a binary tree, return thepreordertraversal of its nodes' values. Example: Input:[1,null,2,3] 1 2 / 3Output:[1,2,3] Follow up:Recursive solution is trivial, could you do it iteratively?
Complete thefunction in the editor below, which hasparameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to thepreOrderfu...