101. Symmetric Tree
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
1 / \ 2 2 / \ / \ 3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1 / \ 2 2 \ \ 3 3
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isSymmetric(TreeNode root) {
if(root==null) return true;
return ismirror(root.left,root.right);
}
public boolean ismirror(TreeNode left,TreeNode right)
{
if(left==null&&right==null) return true;
if(left==null||right==null) return false;
if(left.val!=right.val) return false;
return ismirror(left.left,right.right)&&ismirror(left.right,right.left);
}
}
104. Maximum Depth of Binary Tree
Example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its depth = 3.
public int maxDepth(TreeNode root) {
if(root==null) return 0;
int left = maxDepth(root.left);
int right = maxDepth(root.right);
return left>right?left+1:right+1;
}
用栈深度优先
用两个栈 stack用来深度遍历树 values存储遍历时到每个节点时的高度 max存储最大值
public int maxDepth(TreeNode root) {
if(root==null) return 0;
Stack<TreeNode> stack = new Stack<>();
Stack<Integer> values = new Stack<>();
stack.push(root);
values.push(1);
int max=1;
while(!stack.isEmpty())
{
TreeNode temp = stack.pop();
int val = values.pop();
max = Math.max(max,val);
if(temp.left!=null)
{
stack.push(temp.left);
values.push(val+1);
}
if(temp.right!=null)
{
stack.push(temp.right);
values.push(val+1);
}
}
return max;
}
用队列广度优先遍历
public int maxDepth(TreeNode root) {
if(root==null) return 0;
int count=0;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
int size=0;
TreeNode temp;
while(!queue.isEmpty())
{
size=queue.size();//每一层有多少个元素
while(size-->0)
{
temp=queue.poll();
if(temp.left!=null) queue.offer(temp.left);
if(temp.right!=null) queue.offer(temp.right);
}
count++;//遍历完一层高度+1
}
return count;
}
107. Binary Tree Level Order Traversal II
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> res = new LinkedList<>();
if(root==null) return res;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while(!queue.isEmpty())
{
int size = queue.size();
List<Integer> list = new LinkedList<>();
while(size-->0)
{
TreeNode p = queue.poll();
if(p.left!=null) queue.offer(p.left);
if(p.right!=null) queue.offer(p.right);
list.add(p.val);
}
res.add(0, list);
}
return res;
}
}
方法二
public List<List<Integer>> levelOrderBottom(TreeNode root) {
LinkedList<List<Integer>> res = new LinkedList<>();
if(root==null) return res;
leveladd(res,root,0);
return res;
}
private void leveladd( LinkedList<List<Integer>> res, TreeNode root, int level) {
if(root==null) return;
if(res.size()-1<level) res.addFirst(new LinkedList<Integer>());//如果对应层级的list不存在则新建一个
res.get(res.size()-1-level).add(root.val);//在res中取对应层级的list存储值
leveladd(res,root.left,level+1);
leveladd(res,root.right,level+1);
}
108. Convert Sorted Array to Binary Search Tree
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example:
Given the sorted array: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
0
/ \
-3 9
/ /
-10 5
将排序数组转变为排序二叉树
数组为排序二叉树中序遍历的结果
public TreeNode sortedArrayToBST(int[] nums) {
return convert(nums,0,nums.length-1);
}
private TreeNode convert(int[] nums, int left,int right) {
if(left>right) return null;
TreeNode head = null;
int mid=(left+right)/2;
head = new TreeNode(nums[mid]);
head.left=convert(nums,left,mid-1);
head.right=convert(nums,mid+1,right);
return head;
}