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劍指Offer之Java算法習(xí)題精講二叉樹專項(xiàng)解析

更新時(shí)間：2022年03月18日 10:52:44 作者：明天一定.

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

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 {
    int ans;
    int pre;
    public int getMinimumDifference(TreeNode root) {
        ans = Integer.MAX_VALUE;
        pre = -1;
        method(root);
        return ans;
    }
    public void method(TreeNode root){
        if(root==null) return;
        method(root.left);
        if(pre==-1){
            pre = root.val;
        }else{
            ans = Math.min(ans,root.val-pre);
            pre = root.val;
        }
        method(root.right);
    }
}

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 {
    int ans = 0;
    public int findTilt(TreeNode root) {
        method(root);
        return ans;
    }
    public int method(TreeNode root){
        if(root==null) return 0;
        int l = method(root.left);
        int r = method(root.right);
        ans += Math.abs(l-r);
        return l+r+root.val;
    }
}

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 isSubtree(TreeNode root, TreeNode subRoot) {
        return dfs(root,subRoot);
    }
    public boolean dfs(TreeNode root, TreeNode subRoot){
        if(root==null) return false;
        return cheack(root,subRoot)||dfs(root.left,subRoot)||dfs(root.right,subRoot);
    }
    public boolean cheack(TreeNode root, TreeNode subRoot){
        if(root==null&&subRoot==null) return true;
        if(root==null||subRoot==null||root.val!=subRoot.val) return false;
        return cheack(root.left,subRoot.left)&&cheack(root.right,subRoot.right);
    }
}

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 isSameTree(TreeNode p, TreeNode q) {
        if(p==null&&q==null) return true;
        if(p==null||q==null||q.val!=p.val) return false;
        return isSameTree(p.left,q.left)&&isSameTree(p.right,q.right);
    }
}

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