0110. Balanced Binary Tree (E)
阿新 • • 發佈:2020-12-22
Balanced Binary Tree (E)
題目
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.
Example 1:
Given the following tree [3,9,20,null,null,15,7]
3
/ \
9 20
/ \
15 7
Return true.
Example 2:
Given the following tree [1,2,2,3,3,null,null,4,4]
:
1
/ \
2 2
/ \
3 3
/ \
4 4
Return false.
題意
判斷指定樹是否為平衡二叉樹,即對於樹中任意一個結點,它的左子樹和右子樹高度之差不大於1。
思路
在遞迴求各結點高度的同時,判斷當前結點的左子樹和右子樹高度之差是否大於1。
程式碼實現
Java
class Solution { boolean isBalanced = true; public boolean isBalanced(TreeNode root) { getHeight(root); return isBalanced; } private int getHeight(TreeNode x) { // 空結點時返回高度 // 當已經確定該樹不平衡時,無需再向下遞迴 if (x == null || !isBalanced) { return 0; } int lHeight = getHeight(x.left); int rHeight = getHeight(x.right); if (Math.abs(lHeight - rHeight) > 1) { isBalanced = false; } return Math.max(lHeight, rHeight) + 1; } }
JavaScript
/** * @param {TreeNode} root * @return {boolean} */ var isBalanced = function (root) { return dfs(root) !== -1 } let dfs = function (root) { if (!root) { return 0 } let left = dfs(root.left) let right = dfs(root.right) if (left === -1 || right === -1 || Math.abs(left - right) > 1) { return -1 } return Math.max(left, right) + 1 }