劍指Offer之Java算法習題精講二叉樹與N叉樹

更新時間：2022年03月18日 10:29:24 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 {
    StringBuffer sb = new StringBuffer();
    List<String> list = new ArrayList<String>();
    public List<String> binaryTreePaths(TreeNode root) {
        method(root);
        return list;
    }
    public void method(TreeNode root){
        if(root==null) return;
        int t = sb.length();
        sb.append(root.val);
        if(root.left==null&&root.right==null){
            list.add(sb.toString());
        }
        sb.append("->");
        method(root.left);
        method(root.right);
        sb.delete(t, sb.length());
    }
}

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 sumOfLeftLeaves(TreeNode root) {
        method(root,false);
        return ans;
    }
    public void method(TreeNode root,boolean flag){
        if(root==null) return;
        if(root.left==null&&root.right==null&&flag){
            ans+=root.val;
            return;
        }
        method(root.left,true);
        method(root.right,false);
    }
}

題目三

?解法

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;
    public Node() {}
    public Node(int _val) {
        val = _val;
    }
    public Node(int _val, List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/
 
class Solution {
    public int maxDepth(Node root) {
        if(root==null){
            return 0;
        }
        int maxChildDepth = 0;
        for(int i = 0;i<root.children.size();i++){
            int childDepth = maxDepth(root.children.get(i));
            maxChildDepth = Math.max(maxChildDepth, childDepth);
        }
        return maxChildDepth+1;
    }
}

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