/**
* Source : https://oj.leetcode.com/problems/balanced-binary-tree/
*
*
* Given a binary tree, determine if it is height-balanced.
*
* For this problem, a height-balanced binary tree is defined as a binary tree in which
* the depth of the two subtrees of every node never differ by more than 1.
*/
public class BalancedBinaryTree {
public boolean isBanlanced (TreeNode root) {
return treeDepth(root) == -1 ? false : true;
}
/**
* 判断一棵二叉树是否是高度平衡二叉树
*
* 求出左右子树各自的总高度,如果不是则及早返回-1
*
* @param root
* @return
*/
public int treeDepth (TreeNode root) {
if (root == null) {
return 0;
}
int leftDepth = treeDepth(root.leftChild);
if (leftDepth == -1) {
return -1;
}
int rightDepth = treeDepth(root.rightChild);
if (rightDepth == -1) {
return -1;
}
if (Math.abs(leftDepth-rightDepth) > 1) {
return -1;
}
return Math.max(leftDepth, rightDepth) + 1;
}
public TreeNode createTree (char[] treeArr) {
TreeNode[] tree = new TreeNode[treeArr.length];
for (int i = 0; i < treeArr.length; i++) {
if (treeArr[i] == '#') {
tree[i] = null;
continue;
}
tree[i] = new TreeNode(treeArr[i]-'0');
}
int pos = 0;
for (int i = 0; i < treeArr.length && pos < treeArr.length-1; i++) {
if (tree[i] != null) {
tree[i].leftChild = tree[++pos];
if (pos < treeArr.length-1) {
tree[i].rightChild = tree[++pos];
}
}
}
return tree[0];
}
private class TreeNode {
TreeNode leftChild;
TreeNode rightChild;
int value;
public TreeNode(int value) {
this.value = value;
}
public TreeNode() {
}
}
public static void main(String[] args) {
BalancedBinaryTree balancedBinaryTree = new BalancedBinaryTree();
char[] arr0 = new char[]{'#'};
char[] arr1 = new char[]{'3','9','2','#','#','1','7'};
char[] arr2 = new char[]{'3','9','2','#','#','1','7','5'};
System.out.println(balancedBinaryTree.isBanlanced(balancedBinaryTree.createTree(arr0)));
System.out.println(balancedBinaryTree.isBanlanced(balancedBinaryTree.createTree(arr1)));
System.out.println(balancedBinaryTree.isBanlanced(balancedBinaryTree.createTree(arr2)));
}
}