判断是否为对称二叉树
目录
题目:
分析:
代码:
public boolean isSymmetricChild(TreeNode leftTree,TreeNode rightTree) {
if (leftTree == null && rightTree == null) return true;
//设置两种递归结束条件
if ((leftTree == null && rightTree != null) || (leftTree != null && rightTree == null)) return false;
if (leftTree.val != rightTree.val) return false;
//分别检验对应子树
return isSymmetricChild(leftTree.right,rightTree.left) && isSymmetricChild(leftTree.left,rightTree.right);
}
public boolean isSymmetric(TreeNode root) {
if (root == null) return true;
return isSymmetricChild(root.left,root.right);
}
二叉树的最大深度
题目:
分析:
代码:
public int maxDepth(TreeNode root) {
//设置递归结束条件
if (root == null) return 0;
if (root.left == null && root.right == null) {
return 1;
}
//取最大高度后加上根结点
return Math.max(maxDepth(root.left),maxDepth(root.right))+1;
}
判断一棵二叉树是否是平衡二叉树
题目:
分析:
代码:
private int height(TreeNode root) {
if (root == null) return 0;
//用子问题思路,分别求出所遍历的结点的高度。
int Left = height(root.left);
int Right = height(root.right);
//求出高度立刻判断是否平衡,平衡返回高度,否则返回-1.
if (Left >= 0 && Right >= 0 && Math.abs(Left - Right) <= 1) {
return Math.max(Left,Right) + 1;
}
return -1;
}
public boolean isBalanced(TreeNode root) {
if (root == null) return true;
return height(root) > 0;
}