劍指Offer之Java算法習(xí)題精講二叉樹專題篇下

更新時間：2022年03月18日 10:17:11 作者：明天一定.

跟著思路走，之后從簡單題入手，反復(fù)去看，做過之后可能會忘記，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 {
    public int maxDepth(TreeNode root) {
        return  method(root);
    }
    int method(TreeNode root){
        if(root==null){
            return 0;
        }
        int l = method(root.left);
        int r = method(root.right);
        return Math.max(l, r) + 1;
    }
}

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 diameterOfBinaryTree(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.max(ans,l+r);
        return Math.max(l,r)+1;
    }
}

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 int minDepth(TreeNode root) {
        if(root==null) return 0;
        if(root.left==null&&root.right==null) return 1;
        int min = Integer.MAX_VALUE;
        if(root.left!=null){
            min = Math.min(min,minDepth(root.left));
        }
        if(root.right!=null){
            min = Math.min(min,minDepth(root.right));
        }
        return min+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 {
    List<Integer> list = new ArrayList<Integer>();
    public List<Integer> preorderTraversal(TreeNode root) {
        method(root);
        return list;
    }
    public void method(TreeNode root){
        if(root==null){
            return;
        }
        // 前序
        list.add(root.val);
        method(root.left);
        // 中序
        method(root.right);
        // 后序
    }
}

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