＜03.05＞Leetcode2

左 根 右 

/**
 * 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> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        inorder(root,res);
        return res;
    }
    public void inorder(TreeNode root,List<Integer>res){
        if(root == null)return ;
        inorder(root.left,res);//左
        res.add(root.val);//根
        inorder(root.right,res);//右
    }
}

/**
 * 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) {
        if(root == null)return 0;//终止条件
        else {
            int left = maxDepth(root.left);
            int right = maxDepth(root.right);
            return Math.max(left,right) + 1;
        }
    }
}

226. 翻转二叉树 - 力扣（LeetCode） 

/**
 * 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;//递归结束的条件
        TreeNode temp = root.left;//存下左节点
        root.left = invertTree(root.right);//递归遍历右节点
        root.right = invertTree(temp);//递归更新右节点为左节点
        return root;

    }
}

/**
 * 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 root == null || recur(root.left,root.right);
    }
    public boolean recur(TreeNode L, TreeNode R) {
        if(L == null && R == null)return true;
        if (L == null || R == null || L.val != R.val) return false;
        //要使用L.val和R.val的前提条件是不为null 不能直接L.val == R.val return..
        return recur(L.left, R.right) && recur(L.right, R.left);
    }
}