leetcode刷题（8）二叉树（2）

/**
 * 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) {
    }
}

/**
 * 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) {
    //当根节点为null时我们规定返回true
        if(root == null) {
            return true;
        }
        if(isSymmetricChild(root.left,root.right)) {
            return true;
        }
        return false;
    }
    public boolean isSymmetricChild(TreeNode leftTree,TreeNode rightTree) {
    //判断左右子树结构是否相同
        if(leftTree == null || rightTree == null) {
            if(leftTree == null && rightTree == null) {
                return true;
            }else {
                return false;
            }
        }
        if(leftTree.val != rightTree.val) {
            return false;
        }
        return isSymmetricChild(leftTree.left,rightTree.right)
            && isSymmetricChild(leftTree.right,rightTree.left);
    }
}

/**
 * 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) {
    }
}

3

/**
 * 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) {
    //当root为null时返回0
        if(root == null) {
            return 0;
        }
        int leftDepth = maxDepth(root.left);
        int rightDepth = maxDepth(root.right);
    //返回leftDepth和rightDepth中深度较大的
        return leftDepth>rightDepth?(leftDepth+1):(rightDepth+1);
    }
}

/**
 * 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) {
}

/**
 * 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;
        }
        invertTree(root.left);
        invertTree(root.right);
        TreeNode tmp = root.left;
        root.left = root.right;
        root.right = tmp;
        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 int maxDepth(TreeNode root) {
    //当root为null时返回0，即使刚进来时root为null，也可以考虑到
        if(root == null) {
            return 0;
        }
        int leftTree = maxDepth(root.left);
        //我们在这里直接先进行判断，如果不符合就直接返回，可以减少时间的消耗
        if(leftTree < 0) {
            return -1;
        }
        int rightTree = maxDepth(root.right);
        if(rightTree < 0) {
            return -1;
        }
        if(Math.abs(leftTree-rightTree) <= 1) {
            return Math.max(leftTree,rightTree)+1;
        }else {
            return -1;
        }
    }
    public boolean isBalanced(TreeNode root) {
        if(maxDepth(root) < 0) {
            return false;
        }
        return true;
    }
}