前序Preorder: 先访问根节点,然后访问左子树,最后访问右子树。子树递归同理 中序Inorder: 先访问左子树,然后访问根节点,最后访问右子树. 后序Postorder:先访问左子树,然后访问右子树,最后访问根节点. class Node: def __init__(self, key): self.left = None self.right = None self.val = key def
65publicstaticvoidinOrder(TreeNode root){66if(root ==null)return;67inOrder(root.left);68visit(root);69inOrder(root.right);70}7172publicstaticvoidinOrder2(TreeNode root){73if(root ==null)return;74Stack<TreeNode> stack =newStack<TreeNode>();75while(!stack.empty() || root !=null){76...
1vector<int> preorderTraversal(TreeNode*root) {2vector<int>rVec;3if(!root)4returnrVec;56stack<TreeNode *>st;7st.push(root);8while(!st.empty())9{10TreeNode *tree =st.top();11rVec.push_back(tree->val);12st.pop();13if(tree->right)14st.push(tree->right);15if(tree->left)1...
preorder: root-left-right inorder: left-root-right postorder: left-right-root order指的是root的位置。 recursive算法比较简单,iterative算法比较难想,可是leetcode原题都说了: recursive method is trivial, could you do iteration? 144.Binary Tree Preorder Traversal /*iterative*/ public List<Integer> p...
Given preorder and inorder traversal of a tree, construct the binary tree. 本题就是根据前序遍历和中序遍历的结果还原一个二叉树。 题意很简答,就是递归实现,直接参考代码吧。 查询index部分可以使用Map做查询,建议和下一道题 leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal 中...
Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 思路: 这题的思路与 105 Construct Binary Tree from Preorder and Inorder Traversal 基本相同。
1. Problem Descriptions:Given two integer arrays inorderandpostorderwhereinorderis the inorder traversal of a binary tree andpostorderis the postorder traversal of the same tree, construct and retu…
*/classSolution{public List<Integer>preorderTraversal(TreeNode root){List<Integer>result=newLinkedList<>();TreeNode current=root;TreeNode prev=null;while(current!=null){if(current.left==null){result.add(current.val);current=current.right;}else{// has left, then find the rightmost of left su...
这三种tree traversal一定要记住,属于基本功。自己平时没事就练习一下,有助于打好基础 preorder: root, left, right inorder: left root right postorder: left right root 今天重新做了buildTree 系列,从inorder, postorder中buildTree。 从inorder, preorder 中buildtree。 还有从preorder, postorder中buildtree...
Binary tree traversal: Preorder, Inorder, and Postorder In order to illustrate few of the binary tree traversals, let us consider the below binary tree: Preorder traversal: To traverse a binary tree in Preorder, following operations are carried-out (i) Visit the root, (ii) Traverse the le...