04.04. Check Balance

Description

Implement a function to check if a binary tree is balanced. For the purposes of this question, a balanced tree is defined to be a tree such that the heights of the two subtrees of any node never differ by more than one.

Example 1:


Given tree [3,9,20,null,null,15,7]

    3

   / \

  9  20

    /  \

   15   7

return true.

Example 2:


Given [1,2,2,3,3,null,null,4,4]

      1

     / \

    2   2

   / \

  3   3

 / \

4   4

return false.

Solutions

Solution 1: Recursion (Post-order Traversal)

We design a function $dfs(root)$, which returns the height of the tree with $root$ as the root node. If the tree with $root$ as the root node is balanced, it returns the height of the tree, otherwise, it returns $-1$.

The execution logic of the function $dfs(root)$ is as follows:

If $root$ is null, then return $0$.
Otherwise, we recursively call $dfs(root.left)$ and $dfs(root.right)$, and check whether the return values of $dfs(root.left)$ and $dfs(root.right)$ are $-1$. If not, we check whether $abs(dfs(root.left) - dfs(root.right)) \leq 1$ holds. If it holds, then return $max(dfs(root.left), dfs(root.right)) + 1$, otherwise return $-1$.

In the main function, we only need to call $dfs(root)$, and check whether its return value is $-1$. If not, return true, otherwise return false.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the number of nodes in the binary tree.

Python3JavaC++GoTypeScriptSwift

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution:
    def isBalanced(self, root: TreeNode) -> bool:
        def dfs(root: TreeNode):
            if root is None:
                return 0
            l, r = dfs(root.left), dfs(root.right)
            if l == -1 or r == -1 or abs(l - r) > 1:
                return -1
            return max(l, r) + 1

        return dfs(root) >= 0

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isBalanced(TreeNode root) {
        return dfs(root) >= 0;
    }

    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int l = dfs(root.left);
        int r = dfs(root.right);
        if (l < 0 || r < 0 || Math.abs(l - r) > 1) {
            return -1;
        }
        return Math.max(l, r) + 1;
    }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isBalanced(TreeNode* root) {
        function<int(TreeNode*)> dfs = [&](TreeNode* root) {
            if (!root) {
                return 0;
            }
            int l = dfs(root->left);
            int r = dfs(root->right);
            if (l == -1 || r == -1 || abs(l - r) > 1) {
                return -1;
            }
            return max(l, r) + 1;
        };
        return dfs(root) >= 0;
    }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isBalanced(root *TreeNode) bool {
    var dfs func(*TreeNode) int
    dfs = func(root *TreeNode) int {
        if root == nil {
            return 0
        }
        l, r := dfs(root.Left), dfs(root.Right)
        if l == -1 || r == -1 || abs(l-r) > 1 {
            return -1
        }
        return max(l, r) + 1
    }
    return dfs(root) >= 0
}

func abs(x int) int {
    if x < 0 {
        return -x
    }
    return x
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function isBalanced(root: TreeNode | null): boolean {
    const dfs = (root: TreeNode | null): number => {
        if (!root) {
            return 0;
        }
        const l = dfs(root.left);
        const r = dfs(root.right);
        if (l === -1 || r === -1 || Math.abs(l - r) > 1) {
            return -1;
        }
        return Math.max(l, r) + 1;
    };
    return dfs(root) >= 0;
}

/* class TreeNode {
*    var val: Int
*    var left: TreeNode?
*    var right: TreeNode?
*
*    init(_ val: Int) {
*        self.val = val
*        self.left = nil
*        self.right = nil
*    }
*  }
*/

class Solution {
    func isBalanced(_ root: TreeNode?) -> Bool {
        return dfs(root) >= 0
    }

    private func dfs(_ root: TreeNode?) -> Int {
        guard let root = root else {
            return 0
        }

        let leftHeight = dfs(root.left)
        let rightHeight = dfs(root.right)
        if leftHeight < 0 || rightHeight < 0 || abs(leftHeight - rightHeight) > 1 {
            return -1
        }
        return max(leftHeight, rightHeight) + 1
    }
}

04.04. Check Balance

Description

Solutions

Solution 1: Recursion (Post-order Traversal)

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