二叉樹的深度:遞歸
判斷二叉樹是否是平衡二叉樹:注意二叉樹平衡代表的是所有非葉子節點都是一棵平衡樹 – 而不僅僅是根節點
public class _Q39<T> {
public int TreeDepth(BinaryTreeNode<?> tree){
if(tree == null) return 0;
if(tree.leftChild == null && tree.rightChild == null) return 1;
int left = TreeDepth(tree.leftChild);
int right = TreeDepth(tree.rightChild);
return (left > right ? left : right) + 1;
}
public boolean IsBalance(BinaryTreeNode<?> tree, int depth[]){
if(tree == null) return true;
int depthL[] = {0};
int depthR[] = {0};
// 相當於後序遍歷
if(IsBalance(tree.leftChild, depthL) && IsBalance(tree.rightChild, depthR)){
int diff = depthL[0] - depthR[0];
if(diff <= 1 && diff >= -1){
depth[0] = (depthL[0] > depthR[0] ? depthL[0] : depthR[0]) + 1;
return true;
}
}
return false;
}
}
測試代碼:
public class _Q39Test extends TestCase {
_Q39<?> treeJudge = new _Q39();
public void test(){
BinaryTreeNode<Integer> root = new BinaryTreeNode<>();
BinaryTreeNode<Integer> node1 = new BinaryTreeNode<>();
BinaryTreeNode<Integer> node2 = new BinaryTreeNode<>();
BinaryTreeNode<Integer> node3 = new BinaryTreeNode<>();
BinaryTreeNode<Integer> node4 = new BinaryTreeNode<>();
BinaryTreeNode<Integer> node5 = new BinaryTreeNode<>();
BinaryTreeNode<Integer> node6 = new BinaryTreeNode<>();
root.value = 10;
node1.value = 6;
node2.value = 14;
node3.value = 4;
node4.value = 8;
node5.value = 12;
node6.value = 16;
root.leftChild = node1;
root.rightChild = node2;
//root.rightChild = null;
node1.leftChild = node3;
node1.rightChild = node4;
node2.leftChild = node5;
node2.rightChild = node6;
node3.leftChild = null; node3.rightChild = null;
node4.leftChild = null; node4.rightChild = null;
node5.leftChild = null; node5.rightChild = null;
node6.leftChild = null; node6.rightChild = null;
System.out.println(treeJudge.TreeDepth(root));
int depth[] = {0};
System.out.println(treeJudge.IsBalance(root, depth));
}
}