110. Balanced Binary Tree
Given a binary tree, determine if it is height-balanced.
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 1:
Given the following tree [3,9,20,null,null,15,7]
:
3 / \ 9 20 / \ 15 7
Return true.
方法一 Runtime: 2 ms
/** *求每个节点高度 *如果相等则递归向下计算 * */ public boolean isBalanced(TreeNode root) { if(root==null) return true; int left = getdepth(root.left); int right = getdepth(root.right); if(Math.abs(left-right)>1)//如果高度差>1不平衡 { return false; } return isBalanced(root.left)&&isBalanced(root.right);//继续判断 } public int getdepth(TreeNode node)//求节点高度 { if(node==null) return 0; int left = getdepth(node.left); int right = getdepth(node.right); return left>right ? left +1:right+1; }
方法2 Runtime: 2 ms 需遍历树
class Solution { public boolean flag =true; public boolean isBalanced(TreeNode root) { getdepth(root); return flag; } public int getdepth(TreeNode node) { if(node==null) return 0; int left = getdepth(node.left); int right = getdepth(node.right); if(left-right>1||left-right<-1) { flag = false; } return left>right ? left +1:right+1; } }方法3 在方法2的基础上高改进不必遍历树 什么时候判断不平衡社么时候结束
class Solution { public boolean isBalanced(TreeNode root) { return getdepth(root)!=-1?true:false; } public int getdepth(TreeNode node) { if(node==null) return 0; int left = getdepth(node.left); if(left == -1) return -1; //加入判断 减少递归 int right = getdepth(node.right); if(right == -1) return -1; if(left-right>1||left-right<-1) { return -1; } return left>right ? left +1:right+1; } }
111. Minimum Depth of Binary Tree
Given a binary tree, find its minimum depth.
The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
Note: A leaf is a node with no children.
Example:
Given binary tree [3,9,20,null,null,15,7]
,
3 / \ 9 20 / \ 15 7
return its minimum depth = 2.
求根到叶子节点的最小高度
方法一 按层遍历树 如果某层有叶子节点则返回高度
class Solution { public int minDepth(TreeNode root) { if(root==null) return 0; Queue<TreeNode> queue = new LinkedList<>(); queue.offer(root); int count=1; while(!queue.isEmpty()) { int size = queue.size(); while(size-->0) { TreeNode p = queue.poll(); if(p.left==null&&p.right==null) return count;//判断叶子节点如果是叶子则返回结果 if(p.left!=null) queue.offer(p.left); if(p.right!=null) queue.offer(p.right); } count++; } return count; } }
方法2
class Solution { public int minDepth(TreeNode root) { if(root==null) return 0; int left = minDepth(root.left); int right = minDepth(root.right); // left == 0 right ==0 高=1 // left != 0 right ==0 高=left +0 + 1 // left == 0 right !=0 高=0 +right +1 // left != 0 right !=0 高=Math.min(left,right)+1; return (left==0||right==0)? left+right+1 : Math.min(left,right)+1; } }
112. Path Sum
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
Note: A leaf is a node with no children.
Example:
Given the below binary tree and sum = 22
,
5 / \ 4 8 / / \ 11 13 4 / \ \ 7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2
which sum is 22.
递归版
class Solution { public boolean hasPathSum(TreeNode root, int sum) { if(root==null) return false; int target = sum -root.val; if(root.left==null&&root.right==null) return target==0; return hasPathSum(root.left, target)||hasPathSum(root.right, target); } }
非递归
class Solution { public boolean hasPathSum(TreeNode root, int sum) { if(root==null) return false; Stack<TreeNode> nodestack = new Stack<>();//遍历树 nodestack.push(root); Stack<Integer> sumstack = new Stack<>();//遍历树的时候记录 根到该节点的和 sumstack.push(root.val); int val=0; while(!nodestack.isEmpty()) { TreeNode temp = nodestack.pop(); val=sumstack.pop(); if(temp.left==null&&temp.right==null)//如果为叶子则判断是否为sum { if(val==sum) return true; } if(temp.left!=null) { nodestack.push(temp.left); sumstack.push(temp.left.val+val); } if(temp.right!=null) { nodestack.push(temp.right); sumstack.push(temp.right.val+val); } } return false; } }
118. Pascal's Triangle
In Pascal's triangle, each number is the sum of the two numbers directly above it.
Example:
Input: 5 Output: [ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ]
杨辉三角
class Solution { public List<List<Integer>> generate(int numRows) { List<List<Integer>> res = new ArrayList<>(); if(numRows==0) return res; ArrayList<Integer> row = new ArrayList<>(); row.add(1); res.add(row); List<Integer> temp; for(int i=1;i<numRows;i++)//从第二行开始 { temp=res.get(i-1);//上一行 List<Integer> add = new ArrayList<>(); add.add(1); for(int j=1;j<temp.size();j++) { add.add(temp.get(j-1)+temp.get(j)); //当前行与上一行对应元素的关系 } add.add(1); res.add(add); } return res; } }
119. Pascal's Triangle II
Example:
Input: 3 Output: [1,3,3,1]
获得rowIndex行的杨辉三角结果
方法一 在上题的基础上修改
public List<Integer> getRow(int rowIndex) { List<List<Integer>> res = new ArrayList<>(); ArrayList<Integer> row = new ArrayList<>(); row.add(1); res.add(row); if(rowIndex==0) return res.get(0); List<Integer> temp; for(int i=1;i<=rowIndex;i++)//从第二行开始 { temp=res.get(i-1);//上一行 List<Integer> add = new ArrayList<>(); add.add(1); for(int j=1;j<temp.size();j++) { add.add(temp.get(j-1)+temp.get(j)); //当前行与上一行对应元素的关系 } add.add(1); res.add(add); } return res.get(rowIndex); }
方法2
class Solution { public List<Integer> getRow(int rowIndex) { List<Integer> row = new ArrayList<>(); row.add(1); for(int i=1;i<=rowIndex;i++) { for(int j=0;j<row.size()-1;j++) { row.set(j,row.get(j)+row.get(j+1));//新行1中间的数等于老行对应索引位置的数与后一个数的和 } row.add(0,1); } return row; } }