剑指Offer之Java算法习题精讲二叉树专题篇上

剑指Offer之Java算法习题精讲二叉树专题篇上

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 = righhttp://t;

* }

* }

*/

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==nulIrMgpCiGl) return null;

TreeNode node = new TreeNode(root.val);

node.right = invertTree(root.left);

node.left = invertTree(root.right);

return node;

}

}

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