Leetcode 236. Lowest Common Ancestor of a Binary Tree

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本文详细解析了LeetCode 236题——二叉树的最低公共祖先问题，介绍了通过深度优先遍历及回溯算法找到给定两节点的最低公共祖先的方法。

Leetcode 236. Lowest Common Ancestor of a Binary Tree

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题目描述

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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______3______
       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

输入：树的根root，以及两个树的节点p,q
输出：上述该两个节点的最小公共祖先节点

思路

遍历与回溯：如题目所示，首先对整个树进行深度优先遍历一遍，并记录对应搜索节点所在的层数pl、ql，以及每个节点的前置节点prev。而后根据DFS结果进行回溯。
1. 若pl==ql，若prev[p]==prev[q]则返回prev[q]。否则，搜索节点进行回溯

if(pl==ql){
       if(prev[tp]==prev[tq]){
            return prev[tp];
        }else{
            pl--;
            ql--;
            tp = prev[tp];
            tq = prev[tq];
        }
}

若pl>ql，那么有可能q是p的祖先，否则回溯p；反之亦然。

            }else if(pl>ql){
                if(tq==prev[tp]){
                    return tq;
                }else{
                    pl--;
                    tp = prev[tp];
                }
            }else if(pl<ql){
                if(tp==prev[tq]){
                    return tp;
                }else{
                    ql--;
                    tq = prev[tq];
                }
            }

DFS复杂度 $O(n)$ ,回溯复杂度 $O(n)$

代码

class Solution {
public:
    unordered_map<TreeNode*,TreeNode*> prev;
    int pl=0,ql=0;
    int level = 0;
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        TreeNode* temp = root;
        TreeNode* tp = p;
        TreeNode* tq = q;
        prev[root] = root;
        dfs(temp,p,q);
        while(pl>=0&&ql>=0){
            if(pl==ql){
                if(prev[tp]==prev[tq]){
                    return prev[tp];
                }else{
                    pl--;
                    ql--;
                    tp = prev[tp];
                    tq = prev[tq];
                }
            }else if(pl>ql){
                if(tq==prev[tp]){
                    return tq;
                }else{
                    pl--;
                    tp = prev[tp];
                }
            }else if(pl<ql){
                if(tp==prev[tq]){
                    return tp;
                }else{
                    ql--;
                    tq = prev[tq];
                }
            }

        }

        return root;
    }

    void dfs(TreeNode* root, TreeNode* p, TreeNode* q){
        if(root==NULL||p==NULL||q==NULL){
            return;
        }
        if(root==p){
            pl = level;
        }
        if(root==q){
            ql = level;
        }


        if(root->left!=NULL){
            prev[root->left] = root;
            level++;
            dfs(root->left,p,q);
            level--;
        }

        if(root->right!=NULL){
            prev[root->right] = root;
            level++;
            dfs(root->right,p,q);
            level--;
        }

    }
};