236. Lowest Common Ancestor of a Binary Tree (Medium)

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given the following binary tree:  root = [3,5,1,6,2,0,8,null,null,7,4]

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Note:

• All of the nodes' values will be unique.
• p and q are different and both values will exist in the binary tree.

Solutions

1.

class Solution {
    private TreeNode ans;

    public Solution() {
        // Variable to store LCA node.
        this.ans = null;
    }

    private boolean recurseTree(TreeNode currentNode, TreeNode p, TreeNode q) {
        // If reached the end of a branch, return false.
        if (currentNode == null) {
            return false;
        }

        // Left Recursion. If left recursion find either node, set left = 1 else 0
        int left = this.recurseTree(currentNode.left, p, q) ? 1 : 0;

        // Right Recursion
        int right = this.recurseTree(currentNode.right, p, q) ? 1 : 0;

        // If the current node is one of p or q, or p and q are the same node
        int mid = (currentNode.val == p.val) ? 1 : 0;
        mid += (currentNode.val == q.val) ? 1 : 0;

        // Both nodes are found under node 'currentNode', there are four situations
        // 1. left and right
        // 2. left and mid
        // 3. right and mid
        // 4. mid and mid
        if (mid + left + right >= 2) {
            // Only the lowest parent can set this value, which means only the direct
            // parent can make the sum mid + left + right >= 2.
            // So here we don't have to check if this.ans is null;
            this.ans = currentNode;
        }

        // Return true if any one of the three bool values is True.
        return (mid + left + right > 0);
    }

    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        // Traverse the tree
        this.recurseTree(root, p, q);
        return this.ans;
    }
}

2.

class Solution {

    // This solution does not take one special situation into consideration. That is p
    // and q are the same node.

    private TreeNode ans;

    public Solution() {
        // Variable to store LCA node.
        this.ans = null;
    }

    private boolean recurseTree(TreeNode currentNode, TreeNode p, TreeNode q) {

        // If reached the end of a branch, return false.
        if (currentNode == null) {
            return false;
        }

        // Left Recursion. If left recursion returns true, set left = 1 else 0
        int left = this.recurseTree(currentNode.left, p, q) ? 1 : 0;

        // Right Recursion
        int right = this.recurseTree(currentNode.right, p, q) ? 1 : 0;

        // If the current node is one of p or q
        // It is better to compare the value instead of the object.
        int mid = (currentNode == p || currentNode == q) ? 1 : 0;


        // If any two of the flags left, right or mid become True
        if (mid + left + right >= 2) {
            this.ans = currentNode;
        }

        // Return true if any one of the three bool values is True.
        return (mid + left + right > 0);
    }

    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        // Traverse the tree
        this.recurseTree(root, p, q);
        return this.ans;
    }
}

Incorrect Solutions

References