劍指Offer之Java算法習題精講二叉樹專題篇上

更新時間：2022年03月18日 09:57:45 作者：明天一定.

跟著思路走，之后從簡單題入手，反復去看，做過之后可能會忘記，之后再做一次，記不住就反復做，反復尋求思路和規(guī)律，慢慢積累就會發(fā)現(xiàn)質(zhì)的變化

來和二叉樹玩耍吧~????????????????????????????????

題目一

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);
    }
}

題目二

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;
    }
}

題目三

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;
    }
}

題目四

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);
    }
}

題目五

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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