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...
[LeetCode]94, 144, 145 Binary Tree InOrder, PreOrder, PostOrder Traversal_Medium [LeetCode] 589. N-ary Tree Preorder Traversal_Easy [LeetCode] 590. N-ary Tree Postorder Traversal_Easy [LeetCode] 105. Construct Binary Tree from Preorder and Inorder Traversal_Medium tag: Tree Traversal [Le...
public TreeNode buildTree(int[] preorder, int[] inorder) { if(preorder==null || inorder==null) return null; HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); for(int i=0;i<inorder.length;i++) { map.put(inorder[i],i); } return helper(preorder,0,preorder.l...
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tre...
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。
2、先序方式遍历左子树。 3、先序遍历右子树。 简称:根左右 以下图为例进行先序遍历: image.png 遍历过程如下: [A,A left,A right] -->[A,B,B left, B right,C,C left,C right] --->[A,B,D,E,C,F,F left,G] -->[A,B,D,E,C,F,H,G] ...
Morris traversal: My code: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{publicList<Integer>preorderTraversal(TreeNoderoot){List<Integer>ret=newArrayList<...
题目: Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 根据前序遍历和中序遍历结果构造二叉树。 思路分析: 分析二叉树前序遍历和中序遍历的结果我们发现: 二叉树前序遍历的第一个节点是根节点。 在中序遍历...
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105. Construct Binary Tree from Preorder and Inorder Traversal——tree,程序员大本营,技术文章内容聚合第一站。