标签:flat solution ini || 判断 integer bool search 前序遍历
二叉树的遍历,注意递归和非递归两种思路。
94. Binary Tree Inorder Traversal 二叉树中序遍历
https://leetcode.com/problems/binary-tree-inorder-traversal/
题目:给定二叉树,返回节点值的中序遍历。
思路:
class Solution { public List<Integer> inorderTraversal(TreeNode root) { List<Integer> list = new ArrayList<Integer>(); if(root == null) return list; inShow(list, root); return list; } public void inShow(List<Integer> l, TreeNode node) { if(node.left != null) { inShow(l, node.left); } l.add(node.val); if(node.right != null) { inShow(l, node.right); } } }
100. Same Tree 相同的树
https://leetcode.com/problems/same-tree/
题目:给定两个二叉树,编写一个函数来检查它们是否相同。如果两个二叉树在结构上相同,且节点具有相同的值,则它们被认为是相同的。
思路:
①:判断两棵树是否相同和之前的判断两棵树是否对称都是一样的原理,利用深度优先搜索DFS来递归。
②:还有非递归的解法,因为二叉树的四种遍历(层序,先序,中序,后序)均有各自的迭代和递归的写法,这里我们先来看先序的迭代写法,相当于同时遍历两个数,然后每个节点都进行比较。
class Solution { public boolean isSameTree(TreeNode p, TreeNode q) { if (p == null && q == null) return true; if ((p != null && q == null) || (p == null && q != null) || (p.val != q.val)) return false; return isSameTree(p.left, q.left) && isSameTree(p.right, q.right); } }
129. Sum Root to Leaf Numbers 求根节点到叶节点的值的总和
http://oj.leetcode.com/problems/sum-root-to-leaf-numbers/
题目:如果一个二叉树只包含从0到9的数字,那么每个根到叶的路径都可以表示一个数字.一个例子是根对叶路径1->2->3,它表示数字123。找出所有根到叶数的总和。注意:叶子是一个没有子节点的节点。
思路:
class Solution { int tot = 0; public int sumNumbers(TreeNode root) { if(root==null) return 0; calc(root,0); return tot; } public void calc(TreeNode root, int currsum) { currsum =currsum*10 + root.val; if(root.left==null && root.right==null){ tot+=currsum; return; } if(root.left!=null) calc(root.left,currsum); if(root.right!=null) calc(root.right,currsum); return; } }
144. Binary Tree Preorder Traversal 二叉树前序遍历
https://leetcode.com/problems/binary-tree-preorder-traversal/
题目:给定二叉树,返回其节点值的前序遍历。
思路:
class Solution { // 递归实现 public List<Integer> preorderTraversal(TreeNode root) { List<Integer> list = new ArrayList<Integer>(); if(root == null) return list; frontShow(list, root); return list; } public void frontShow(List<Integer> l, TreeNode node) { l.add(node.val); if(node.left != null) { frontShow(l, node.left); } if(node.right != null) { frontShow(l, node.right); } } }
class Solution { // 非递归实现 public List<Integer> preorderTraversal(TreeNode root) { List<Integer> list = new LinkedList<Integer>(); Stack<TreeNode> rights = new Stack<TreeNode>(); while(root != null) { list.add(root.val); if(root.right != null) { rights.push(root.right); } root = root.left; if(root == null && !rights.isEmpty()) { root = rights.pop(); } } return list; } }
145. Binary Tree Postorder Traversal 二叉树后序遍历
https://leetcode.com/problems/binary-tree-postorder-traversal/
题目:给定二叉树,返回节点值的后序遍历。
思路:
class Solution { // 递归实现 public List<Integer> postorderTraversal(TreeNode root) { List<Integer> list = new ArrayList<Integer>(); if(root == null) return list; backShow(list, root); return list; } public void backShow(List<Integer> l, TreeNode node) { if(node.left != null) { backShow(l, node.left); } if(node.right != null) { backShow(l, node.right); } l.add(node.val); } }
// TODO
遍历变种:http://oj.leetcode.com/problems/path-sum/
遍历变种:http://oj.leetcode.com/problems/path-sum-ii/
遍历变种:http://oj.leetcode.com/problems/maximum-depth-of-binary-tree/
遍历变种:http://oj.leetcode.com/problems/minimum-depth-of-binary-tree/
重建二叉树:http://oj.leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
重建二叉树:http://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/
层次遍历变种:http://oj.leetcode.com/problems/binary-tree-zigzag-level-order-traversal/
遍历变种:http://oj.leetcode.com/problems/symmetric-tree/
遍历应用:http://oj.leetcode.com/problems/binary-tree-maximum-path-sum/
遍历应用:http://oj.leetcode.com/problems/balanced-binary-tree/
遍历应用:http://oj.leetcode.com/problems/recover-binary-search-tree/
遍历应用:http://oj.leetcode.com/problems/flatten-binary-tree-to-linked-list/
level遍历:http://oj.leetcode.com/problems/binary-tree-level-order-traversal/
level 遍历:http://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/
level 遍历变种:http://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/
level 遍历变种:http://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/
标签:flat solution ini || 判断 integer bool search 前序遍历
原文地址:https://www.cnblogs.com/nomad1c/p/11576628.html