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...
confused what"{1,#,2,3}"means?> read more on how binary tree is serialized on OJ. 思路: preorder用栈两三下就写完了 1vector<int> preorderTraversal(TreeNode *root) {2vector<int>nodes;3if(root == NULL)returnnodes;4stack<TreeNode *>tStack;5tStack.push(root);6while(!tStack.empty()...
class Solution(object): def _preorderTraversal(self, root, result): if root: result.append(root.val) self._preorderTraversal(root.left, result) self._preorderTraversal(root.right, result) def preorderTraversal(self, root): """ :type root: TreeNode :rtype: List[int] """ if root ==...
while(inorder[inorderLeftTreeEnd] != preorder[preBeg]) ++inorderLeftTreeEnd; int leftTreeSize = inorderLeftTreeEnd - inBeg; int rightTreeSize = (inBeg + size) - inorderLeftTreeEnd - 1; curNode->left = buildTree(preorder, preBeg+1, inorder, inBeg, leftTreeSize); curNode->righ...
来自专栏 · LeetCode刷题 Construct Binary Tree from Preorder and Inorder Traversal 题目描述(中等难度) 根据二叉树的先序遍历和中序遍历还原二叉树。 解法一 递归 先序遍历的顺序是根节点,左子树,右子树。中序遍历的顺序是左子树,根节点,右子树。
public List<Integer> preorderTraversal(TreeNode root) { List<Integer> list = new ArrayList<>(); preorderTraversalHelper(root, list); return list; } private void preorderTraversalHelper(TreeNode root, List<Integer> list) { if (root == null) { return; } list.add(root.val); preorderTrav...
总结: pre order tree ** Anyway, Good luck, Richardo! My code: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * }
Pre-Order Traversal 三种解法: Recursive Iterate 用Stack Morris Traversal (好处是,Pre-Order和In-Order代码只有很小的改动) Morris Pre-Order Traversal (LeetCode 144) (Medium) 1...If left child is null, print the current node data. Move to right child. ...
*/classSolution{public:vector<int>postorderTraversal(TreeNode*root){vector<int>v;if(NULL==root)returnv;stack<TreeNode*>s;s.push(root);TreeNode*pre=root;while(!s.empty()){TreeNode*cur=s.top();if((NULL==cur->left&&NULL==cur->right)||(pre==cur->left||pre==cur->right)){s.pop...
Can you solve this real interview question? Verify Preorder Serialization of a Binary Tree - One way to serialize a binary tree is to use preorder traversal. When we encounter a non-null node, we record the node's value. If it is a null node, we record u