剑指Offer之Java算法习题精讲二叉树专题篇上
作者：明天一定.
跟着思路走，之后从简单题入手，反复去看，做过之后可能会忘记，之后再做一次，记不住就反复做，反复寻求思路和规律，慢慢积累就会发现质的变化
来和二叉树玩耍吧~💕💕💕💕💕💕💕💕💕💕💕💕💕💕💕💕
题目一
解法
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean isSymmetric(TreeNode root) {
return method(root.left,root.right);
}
public boolean method(TreeNode l,TreeNode r){
if(l==null&&r==null) return true;
if(l==null||r==null||l.val!=r.val) return false;
return method(l.left,r.right)&&method(l.right,r.left);
}
}题目二
解法
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode sortedArrayToBST(int[] nums) {
return method(nums,0,nums.length-1);
}
public TreeNode method(int[] nums,int l,int r){
if(l>r) return null;
int mid = l+(r-l)/2;
TreeNode root = new TreeNode(nums[mid]);
root.left = method(nums,l,mid-1);
root.right = method(nums,mid+1,r);
return root;
}
}题目三
解法
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean isBalanced(TreeNode root) {
if(root==null) return true;
return Math.abs(method(root.left)-method(root.right))<=1&&isBalanced(root.left)&&isBalanced(root.right);
}
public int method(TreeNode root){
if(root==null) return 0;
return Math.max(method(root.left),method(root.right))+1;
}
}题目四
解法
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean hasPathSum(TreeNode root, int targetSum) {
if(root==null) return false;
if(root.left == null && root.right == null) return targetSum==root.val;
return hasPathSum(root.left,targetSum-root.val)||hasPathSum(root.right,targetSum-root.val);
}
}题目五
解法
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode invertTree(TreeNode root) {
if(root==null) return null;
TreeNode node = new TreeNode(root.val);
node.right = invertTree(root.left);
node.left = invertTree(root.right);
return node;
}
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