力扣labuladong——一刷day30

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从遍历角度和分解角度处理树问题

前言

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一、力扣104. 二叉树的最大深度

中序遍历

/**
 * 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 depth = 0, res = 0;
    public int maxDepth(TreeNode root) {
        traverse(root);
        return res;
    }
    public void traverse(TreeNode root){
        if(root == null){
            return ;
        }
        depth ++;
        traverse(root.left);
        res = Math.max(depth,res);
        traverse(root.right);
        depth --;
    }
}

前序遍历

/**
 * 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 depth = 0, res = 0;
    public int maxDepth(TreeNode root) {
        traverse(root);
        return res;
    }
    public void traverse(TreeNode root){
        if(root == null){
            return ;
        }
        depth ++;
        res = Math.max(depth,res);
        traverse(root.left);
        traverse(root.right);
        depth --;
    }
}

分解

/**
 * 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 depth(root);
    }
    public int depth(TreeNode root){
        if(root ==  null){
            return 0;
        }
        int leftDepth = depth(root.left);
        int rightDepth = depth(root.right);
        return leftDepth > rightDepth ? leftDepth+1 : rightDepth + 1;
    }
}

二、力扣543. 二叉树的直径

/**
 * 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 len = 0;
    public int diameterOfBinaryTree(TreeNode root) {
        fun(root);
        return len;
    }
    public int fun(TreeNode root){
        if(root == null){
            return 0;
        }
        int l = fun(root.left);
        int r = fun(root.right);
        int my = l+r;
        len = Math.max(len,my);
        return Math.max(l,r) + 1;
    }
}

三、力扣144. 二叉树的前序遍历

/**
 * 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 List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> list = new LinkedList<>();
        if(root == null){
            return list;
        }
        list.add(root.val);
        list.addAll(preorderTraversal(root.left));
        list.addAll(preorderTraversal(root.right));
        return list;
    }
}